Chapter 14 Factorisation (Additional Questions)

Given:

The algebraic expression is $12x^2y$.

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To Find:

The factors of the given algebraic expression.

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Solution:

To find the factors of an algebraic expression, we need to consider the factors of the numerical coefficient and the factors involving the variables.

The numerical coefficient is 12.

We can find the prime factors of 12:

So, $12 = 2^2 \times 3^1$.

The integer factors of 12 are 1, 2, 3, 4, 6, and 12.

The variable part of the expression is $x^2y$.

The factors involving the variables are $1, x, x^2, y, xy, x^2y$.

The factors of the algebraic expression $12x^2y$ are all possible products formed by taking any integer factor of 12 and any variable factor of $x^2y$.

These factors include:

Individual numerical factors: $1, 2, 3, 4, 6, 12$

Individual variable factors: $x, x^2, y$

Combinations (products) of numerical and variable factors, such as $2x, 3y, 4x^2, 6xy, 12x^2y$, etc.

Essentially, the set of all factors of $12x^2y$ consists of 1, all integer factors of 12, all variable factors formed from $x^2$ and $y$, and all products of these integer and variable factors.

Let's examine the given options:

(A) $12, x, y$ - This is an incomplete set of factors.

(B) $2, 6, x^2, y$ - This is also an incomplete set of factors.

(C) $1, 2, 3, 4, 6, 12, x, x^2, y, xy, x^2y, ...$ (all possible integer and variable factors) - This option correctly describes the entire set of factors, acknowledging that the list is extensive and includes all combinations.

Option (C) is the correct answer as it describes the complete set of factors.

(D) $1, 12, x^2, y$ - This is an incomplete set of factors.

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Answer:

The correct option is (C).

Given:

The two algebraic expressions are $6ab$ and $9bc$.

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To Find:

The common factors of the given algebraic expressions.

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Solution:

To find the common factors of two algebraic expressions, we find the factors of each expression and then identify the factors that are common to both.

Consider the first expression, $6ab$.

The numerical coefficient is 6. Let's find its prime factors:

So, $6 = 2 \times 3$. The integer factors of 6 are 1, 2, 3, 6.

The variable part is $ab$. The variable factors are $a, b, ab$.

The factors of $6ab$ include 1, all integer factors of 6, all variable factors of $ab$, and all possible products of these factors. The factors are 1, 2, 3, 6, $a, b, ab, 2a, 2b, 2ab, 3a, 3b, 3ab, 6a, 6b, 6ab$.

Consider the second expression, $9bc$.

The numerical coefficient is 9. Let's find its prime factors:

So, $9 = 3 \times 3 = 3^2$. The integer factors of 9 are 1, 3, 9.

The variable part is $bc$. The variable factors are $b, c, bc$.

The factors of $9bc$ include 1, all integer factors of 9, all variable factors of $bc$, and all possible products of these factors. The factors are 1, 3, 9, $b, c, bc, 3b, 3c, 3bc, 9b, 9c, 9bc$.

Now, let's find the factors that are common to both $6ab$ and $9bc$.

Common numerical factors of 6 and 9: We look at the integer factors: 1, 2, 3, 6 (for 6) and 1, 3, 9 (for 9). The common integer factors are 1 and 3.

Common variable factors of $ab$ and $bc$: We look at the variable factors: $a, b, ab$ (for $ab$) and $b, c, bc$ (for $bc$). The common variable factor is $b$. (Note that 1 is also a common variable factor, representing no variable).

The common factors of $6ab$ and $9bc$ are formed by taking the product of common numerical factors and common variable factors.

Possible combinations:

$1 \times 1 = 1$

$1 \times b = b$

$3 \times 1 = 3$

$3 \times b = 3b$

Thus, the common factors are 1, 3, $b$, and $3b$.

Now let's compare this list with the given options:

(A) $3, b$ - This lists 3 and $b$, which are common factors, but not all of the common factors listed in the options.

(B) $3b$ - This lists $3b$, which is a common factor, but not all of the common factors listed in the options.

(C) $3, b, 3b$ - This lists 3, $b$, and $3b$. These are all common factors. Compared to options (A) and (B), this option provides a more complete list of the non-trivial common factors (excluding 1).

(D) $a, c$ - Neither $a$ nor $c$ are common factors of $6ab$ and $9bc$.

Based on the options provided, option (C) lists the set of common factors excluding 1, which is a standard way of presenting common factors in multiple-choice questions.

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

The algebraic expression is $18x^3y^2 - 27x^2y^3$.

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To Find:

Factorise the given algebraic expression by taking out the common factor.

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Solution:

To factorise the expression $18x^3y^2 - 27x^2y^3$ by taking out the common factor, we need to find the Greatest Common Factor (GCF) of the two terms $18x^3y^2$ and $27x^2y^3$.

First, let's find the GCF of the numerical coefficients, which are 18 and 27.

Prime factorization of 18:

So, $18 = 2 \times 3^2$.

Prime factorization of 27:

So, $27 = 3^3$.

The common prime factors are 3. The lowest power of 3 common to both is $3^2 = 9$.

The GCF of 18 and 27 is 9.

Next, let's find the GCF of the variable parts.

For the variable $x$: The powers are $x^3$ and $x^2$. The lowest power is $x^2$. The common factor is $x^2$.

For the variable $y$: The powers are $y^2$ and $y^3$. The lowest power is $y^2$. The common factor is $y^2$.

The GCF of the entire expression is the product of the GCF of the numerical coefficients and the GCF of the variable parts.

GCF $= 9 \times x^2 \times y^2 = 9x^2y^2$.

Now, we factor out the GCF from each term:

$18x^3y^2 = 9x^2y^2 \times (2x)$

$-27x^2y^3 = 9x^2y^2 \times (-3y)$

So, $18x^3y^2 - 27x^2y^3 = 9x^2y^2(2x - 3y)$.

Comparing this with the given options:

(A) $9xy(2x^2y - 3xy^2)$

(B) $9x^2y^2(2x - 3y)$

(C) $9x^3y^3(2 - 3)$

(D) $3x^2y^2(6x - 9y)$

Our factored expression matches option (B).

Option (B) is the correct answer.

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Answer:

The correct option is (B).

Given:

The algebraic expression is $ax - ay + bx - by$.

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To Find:

Factorise the given expression by grouping the terms.

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Solution:

To factorise the expression $ax - ay + bx - by$ by grouping, we group terms that have common factors.

Group the first two terms and the last two terms:

$(ax - ay) + (bx - by)$

Now, factor out the common factor from each group.

In the first group, $(ax - ay)$, the common factor is $a$.

$ax - ay = a(x - y)$

In the second group, $(bx - by)$, the common factor is $b$.

$bx - by = b(x - y)$

Substitute these factored forms back into the grouped expression:

$a(x - y) + b(x - y)$

Now, observe that the term $(x - y)$ is common to both parts of the expression.

Factor out the common binomial factor $(x - y)$:

$(x - y)(a + b)$

The factored form of the expression is $(a+b)(x-y)$.

Comparing this with the given options:

(A) $(a+b)(x-y)$ - This matches our result.

(B) $(a-b)(x+y)$

(C) $(a+x)(b-y)$

(D) $(a-x)(b+y)$

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The algebraic expression is $49m^2 - 36n^2$.

The identity to use is $a^2 - b^2 = (a+b)(a-b)$.

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To Find:

Factorise the given expression using the specified identity.

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Solution:

We are asked to factorise the expression $49m^2 - 36n^2$ using the identity $a^2 - b^2 = (a+b)(a-b)$.

First, we need to express each term in the form of a perfect square.

The first term is $49m^2$. We can write $49$ as $7^2$ and $m^2$ as $(m)^2$. So, $49m^2 = 7^2 m^2 = (7m)^2$.

The second term is $36n^2$. We can write $36$ as $6^2$ and $n^2$ as $(n)^2$. So, $36n^2 = 6^2 n^2 = (6n)^2$.

Now, the expression $49m^2 - 36n^2$ can be written as $(7m)^2 - (6n)^2$.

This expression is in the form $a^2 - b^2$, where $a = 7m$ and $b = 6n$.

Using the identity $a^2 - b^2 = (a+b)(a-b)$, we substitute the values of $a$ and $b$:

$(7m)^2 - (6n)^2 = (7m + 6n)(7m - 6n)$.

Thus, the factored form of $49m^2 - 36n^2$ is $(7m + 6n)(7m - 6n)$.

Comparing this with the given options:

(A) $(7m - 6n)(7m - 6n)$

(B) $(7m + 6n)(7m + 6n)$

(C) $(7m + 6n)(7m - 6n)$ - This matches our result.

(D) $(49m + 36n)(49m - 36n)$

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

The algebraic expression is $p^2 + 10p + 25$.

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To Find:

Factorise the given expression using a suitable identity.

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Solution:

We are asked to factorise the expression $p^2 + 10p + 25$ using a suitable identity.

The given expression is a trinomial ($p^2$, $10p$, and $25$). We can check if it fits the form of a perfect square trinomial, which is $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$.

The first term is $p^2$, which is the square of $p$. So we can consider $a = p$.

The last term is $25$, which is the square of $5$. So we can consider $b = 5$.

Now let's check the middle term, $10p$, against the form $2ab$.

If $a=p$ and $b=5$, then $2ab = 2 \times p \times 5 = 10p$.

Since the middle term matches $2ab$ and the first and last terms are perfect squares of $a$ and $b$ respectively, the expression $p^2 + 10p + 25$ fits the identity:

$a^2 + 2ab + b^2 = (a+b)^2$

Substituting $a=p$ and $b=5$ into the identity:

$p^2 + 2(p)(5) + 5^2 = (p+5)^2$

$p^2 + 10p + 25 = (p+5)^2$

Thus, the factored form of the expression is $(p+5)^2$.

Comparing this with the given options:

(A) $(p+5)^2$ - This matches our result.

(B) $(p-5)^2$

(C) $(p+10)(p+15)$

(D) $(p+5)(p-5)$

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The algebraic expression is $16y^2 - 40y + 25$.

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To Find:

Factorise the given expression using a suitable identity.

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Solution:

We are asked to factorise the expression $16y^2 - 40y + 25$ using a suitable identity.

The given expression is a trinomial. We can check if it is in the form of a perfect square trinomial, either $a^2 + 2ab + b^2 = (a+b)^2$ or $a^2 - 2ab + b^2 = (a-b)^2$.

Let's examine the terms:

The first term is $16y^2$. We can write this as $(4y)^2$. So, we can let $a = 4y$.

The last term is $25$. We can write this as $5^2$. So, we can let $b = 5$.

Now, let's check the middle term, which is $-40y$, against the forms $2ab$ or $-2ab$.

If we use the form $a^2 - 2ab + b^2$, the middle term should be $-2ab$.

Calculate $-2ab$ with $a=4y$ and $b=5$:

$-2ab = -2 \times (4y) \times 5 = -40y$.

The calculated middle term, $-40y$, matches the middle term of the given expression.

Therefore, the expression $16y^2 - 40y + 25$ is a perfect square trinomial of the form $a^2 - 2ab + b^2$, where $a = 4y$ and $b = 5$.

Using the identity $a^2 - 2ab + b^2 = (a-b)^2$, we substitute $a=4y$ and $b=5$:

$16y^2 - 40y + 25 = (4y)^2 - 2(4y)(5) + 5^2 = (4y - 5)^2$.

Thus, the factored form of $16y^2 - 40y + 25$ is $(4y - 5)^2$.

Comparing this result with the given options:

(A) $(4y+5)^2 = (4y)^2 + 2(4y)(5) + 5^2 = 16y^2 + 40y + 25$ (Incorrect)

(B) $(4y-5)^2 = (4y)^2 - 2(4y)(5) + 5^2 = 16y^2 - 40y + 25$ (Correct)

(C) $(16y-25)(y-1) = 16y^2 - 16y - 25y + 25 = 16y^2 - 41y + 25$ (Incorrect)

(D) $(4y-5)(4y+5) = (4y)^2 - 5^2 = 16y^2 - 25$ (Incorrect)

Option (B) is the correct answer.

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Answer:

The correct option is (B).

Given:

The trinomial is $x^2 + 7x + 12$.

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To Find:

Factorise the trinomial $x^2 + 7x + 12$.

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Solution:

To factorise a trinomial of the form $x^2 + bx + c$, we need to find two numbers, let's say $p$ and $q$, such that their sum is equal to the coefficient of $x$ (which is $b$) and their product is equal to the constant term (which is $c$).

In the given trinomial $x^2 + 7x + 12$, we have $b=7$ and $c=12$.

We need to find two numbers $p$ and $q$ such that:

$p + q = 7$

$p \times q = 12$

Let's list pairs of integers whose product is 12 and check their sums:

Pairs that multiply to 12 are (1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4).

Let's calculate the sum for each pair:

$1 + 12 = 13$

$2 + 6 = 8$

$3 + 4 = 7$

$-1 + (-12) = -13$

$-2 + (-6) = -8$

$-3 + (-4) = -7$

The pair of numbers that sums to 7 and multiplies to 12 is 3 and 4.

So, we can rewrite the middle term $7x$ as the sum of $3x$ and $4x$.

$x^2 + 7x + 12 = x^2 + 3x + 4x + 12$

Now, we can factor by grouping the terms:

$x^2 + 3x + 4x + 12 = (x^2 + 3x) + (4x + 12)$

Factor out the common factor from the first group $(x^2 + 3x)$: The common factor is $x$.

$x(x + 3)$

Factor out the common factor from the second group $(4x + 12)$: The common factor is 4.

$4(x + 3)$

Now, the expression becomes:

$x(x + 3) + 4(x + 3)$

Notice that $(x + 3)$ is a common binomial factor in both terms. Factor out $(x + 3)$.

$(x + 3)(x + 4)$

Thus, the factored form of the trinomial $x^2 + 7x + 12$ is $(x+3)(x+4)$.

Comparing our result with the given options:

(A) $(x+3)(x+4)$ - Matches our factored form.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The trinomial is $m^2 - 11m + 24$.

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To Find:

Factorise the trinomial $m^2 - 11m + 24$.

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Solution:

To factorise a trinomial of the form $x^2 + bx + c$, we need to find two numbers, say $p$ and $q$, such that their sum is equal to the coefficient of $x$ (which is $b$) and their product is equal to the constant term (which is $c$).

In the given trinomial $m^2 - 11m + 24$, the variable is $m$. We have $b = -11$ and $c = 24$.

We need to find two numbers $p$ and $q$ such that:

$p + q = -11$

$p \times q = 24$

Since the product $p \times q$ is positive (24) and the sum $p+q$ is negative (-11), both numbers $p$ and $q$ must be negative.

Let's consider pairs of negative integers whose product is 24 and check their sums:

Pairs of negative integers that multiply to 24 are: (-1, -24), (-2, -12), (-3, -8), (-4, -6).

Let's find the sum for each pair:

$-1 + (-24) = -25$

$-2 + (-12) = -14$

$-3 + (-8) = -11$

$-4 + (-6) = -10$

The pair of numbers that satisfies both conditions (sum is -11 and product is 24) is -3 and -8.

Now, we can rewrite the middle term $-11m$ as the sum of $-3m$ and $-8m$:

$m^2 - 11m + 24 = m^2 - 3m - 8m + 24$

Next, we factor by grouping the terms:

$(m^2 - 3m) + (-8m + 24)$

Factor out the common factor from the first group $(m^2 - 3m)$: The common factor is $m$.

$m(m - 3)$

Factor out the common factor from the second group $(-8m + 24)$: The common factor is -8.

$-8(m - 3)$

The expression now becomes:

$m(m - 3) - 8(m - 3)$

Notice that the binomial $(m - 3)$ is common to both terms. Factor out $(m - 3)$.

$(m - 3)(m - 8)$

Thus, the factored form of the trinomial $m^2 - 11m + 24$ is $(m-3)(m-8)$.

Let's compare this result with the given options:

(A) $(m-3)(m-8)$ - This matches our factored form.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The expression to be divided is $20x^4$ by $5x^2$. This can be written as $\frac{20x^4}{5x^2}$.

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To Find:

The result of the division $\frac{20x^4}{5x^2}$.

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Solution:

To divide $20x^4$ by $5x^2$, we can divide the numerical coefficients and the variable parts separately.

Divide the numerical coefficients:

$\frac{20}{5} = 4$

Divide the variable parts using the rule of exponents, which states that for any non-zero base $a$ and integers $m$ and $n$, $\frac{a^m}{a^n} = a^{m-n}$.

For the variable part $x^4$ divided by $x^2$, we have $m=4$ and $n=2$:

$\frac{x^4}{x^2} = x^{4-2} = x^2$

Now, multiply the results from the numerical and variable divisions:

$4 \times x^2 = 4x^2$

So, the result of dividing $20x^4$ by $5x^2$ is $4x^2$.

Comparing this with the given options:

(A) $4x^6$

(B) $4x^2$ - This matches our result.

(C) $15x^2$

(D) $4x^{-2}$

Option (B) is the correct answer.

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Answer:

The correct option is (B).

Given:

The expression to be divided is $(16y^3 - 8y^2 + 4y)$ by $4y$.

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To Find:

The result of the division $(16y^3 - 8y^2 + 4y) \div 4y$.

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Solution:

To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial divisor.

The expression is $\frac{16y^3 - 8y^2 + 4y}{4y}$.

We can rewrite this as the sum of individual fractions:

$\frac{16y^3}{4y} - \frac{8y^2}{4y} + \frac{4y}{4y}$

Now, perform each division separately:

First term: $\frac{16y^3}{4y}$

Divide the numerical coefficients: $\frac{16}{4} = 4$.

Divide the variable parts using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$: $\frac{y^3}{y^1} = y^{3-1} = y^2$.

So, $\frac{16y^3}{4y} = 4y^2$.

Second term: $-\frac{8y^2}{4y}$

Divide the numerical coefficients: $\frac{-8}{4} = -2$.

Divide the variable parts: $\frac{y^2}{y^1} = y^{2-1} = y^1 = y$.

So, $-\frac{8y^2}{4y} = -2y$.

Third term: $\frac{4y}{4y}$

Divide the numerical coefficients: $\frac{4}{4} = 1$.

Divide the variable parts: $\frac{y^1}{y^1} = y^{1-1} = y^0 = 1$.

So, $\frac{4y}{4y} = 1$.

Combine the results of the individual divisions:

$4y^2 - 2y + 1$

Thus, $(16y^3 - 8y^2 + 4y) \div 4y = 4y^2 - 2y + 1$.

Compare this result with the given options:

(A) $4y^2 - 2y + 1$ - This matches our result.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The expression to be divided is $(z^2 - 9)$ by $(z - 3)$.

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To Find:

The result of the division $(z^2 - 9) \div (z - 3)$.

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Solution:

We need to compute $\frac{z^2 - 9}{z - 3}$.

Observe that the numerator, $z^2 - 9$, is a difference of squares. We can write 9 as $3^2$.

So, $z^2 - 9 = z^2 - 3^2$.

Using the identity for the difference of squares, $a^2 - b^2 = (a - b)(a + b)$, with $a = z$ and $b = 3$, we can factor the numerator:

$z^2 - 3^2 = (z - 3)(z + 3)$

Now substitute the factored form of the numerator back into the division expression:

$\frac{z^2 - 9}{z - 3} = \frac{(z - 3)(z + 3)}{z - 3}$

Assuming $z - 3 \neq 0$, we can cancel out the common factor $(z - 3)$ from the numerator and the denominator:

$\frac{\cancel{(z - 3)}(z + 3)}{\cancel{(z - 3)}} = z + 3$

Thus, the result of dividing $(z^2 - 9)$ by $(z - 3)$ is $z + 3$.

Compare this result with the given options:

(A) $z - 3$

(B) $z + 3$ - This matches our result.

(C) $z^2$

(D) $z - 9$

Option (B) is the correct answer.

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Answer:

The correct option is (B).

Given:

The trinomial to be factorised is $6x^2 + 5x - 6$.

Potential factorisations are provided in the options, along with their expansions.

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To Find:

The correct factorisation of the trinomial $6x^2 + 5x - 6$ from the given options.

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Solution:

We are given four options, each providing a potential factorisation and its expanded form. To find the correct factorisation, we need to check which expanded form matches the original trinomial $6x^2 + 5x - 6$.

Let's examine the expansion provided for each option:

Option (A): $(2x+3)(3x-2)$

As given in the option:

$(2x+3)(3x-2) = 6x^2 - 4x + 9x - 6$

Combine the like terms $(-4x + 9x)$:

$6x^2 + (9-4)x - 6$

$6x^2 + 5x - 6$

This expanded form matches the given trinomial $6x^2 + 5x - 6$.

Let's briefly look at the other options as provided:

Option (B): $(2x-3)(3x+2) = 6x^2 + 4x - 9x - 6 = 6x^2 - 5x - 6$ (Does not match)

Option (C): $(6x+1)(x-6) = 6x^2 - 36x + x - 6 = 6x^2 - 35x - 6$ (Does not match)

Option (D): $(6x-1)(x+6) = 6x^2 + 36x - x - 6 = 6x^2 + 35x - 6$ (Does not match)

Since the expansion of $(2x+3)(3x-2)$ yields $6x^2 + 5x - 6$, option (A) is the correct factorisation.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The algebraic expression is $a^4 - b^4$.

Four potential factorisations are given in the options.

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To Find:

Which of the given options is NOT a factorisation of $a^4 - b^4$.

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Solution:

We need to factorise the expression $a^4 - b^4$ and compare the result(s) with the given options. We can also expand each option to see if it equals $a^4 - b^4$.

The expression $a^4 - b^4$ can be written as a difference of squares:

$a^4 - b^4 = (a^2)^2 - (b^2)^2$

Using the identity $X^2 - Y^2 = (X - Y)(X + Y)$, where $X = a^2$ and $Y = b^2$, we get:

$a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$

This matches Option (A). So, Option (A) is a factorisation.

Now, observe that the term $(a^2 - b^2)$ in the expression from Option (A) is also a difference of squares, $a^2 - b^2 = (a - b)(a + b)$.

Substituting this into the result of Option (A):

$a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) = (a - b)(a + b)(a^2 + b^2)$

This matches Option (B). So, Option (B) is also a factorisation.

Now let's consider Option (C): $(a - b)^2 (a + b)^2$.

$(a - b)^2 (a + b)^2 = [(a - b)(a + b)]^2$

Using the difference of squares identity inside the brackets, $(a - b)(a + b) = a^2 - b^2$:

$[(a - b)(a + b)]^2 = (a^2 - b^2)^2$

Expanding this square using the identity $(X - Y)^2 = X^2 - 2XY + Y^2$, where $X = a^2$ and $Y = b^2$:

$(a^2 - b^2)^2 = (a^2)^2 - 2(a^2)(b^2) + (b^2)^2 = a^4 - 2a^2b^2 + b^4$

Comparing $a^4 - 2a^2b^2 + b^4$ with the original expression $a^4 - b^4$, we see they are not equal (unless $a=0$ or $b=0$).

Therefore, Option (C) is NOT a factorisation of $a^4 - b^4$.

Let's verify Option (D) by expanding it:

Option (D): $(a - b)(a^3 + a^2b + ab^2 + b^3)$.

Multiply each term in the first bracket by each term in the second bracket:

$(a - b)(a^3 + a^2b + ab^2 + b^3) = a(a^3 + a^2b + ab^2 + b^3) - b(a^3 + a^2b + ab^2 + b^3)$

$= (a \cdot a^3 + a \cdot a^2b + a \cdot ab^2 + a \cdot b^3) + (-b \cdot a^3 -b \cdot a^2b - b \cdot ab^2 - b \cdot b^3)$

$= a^4 + a^3b + a^2b^2 + ab^3 - a^3b - a^2b^2 - ab^3 - b^4$

Group and cancel out the like terms:

$= a^4 + (a^3b - a^3b) + (a^2b^2 - a^2b^2) + (ab^3 - ab^3) - b^4$

$= a^4 + 0 + 0 + 0 - b^4$

$= a^4 - b^4$

This expansion equals the original expression $a^4 - b^4$. Therefore, Option (D) is a factorisation.

We have found that Options (A), (B), and (D) are valid factorisations of $a^4 - b^4$, while Option (C) is not.

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Answer:

The correct option is (C).

Given:

Assertion (A): The expression $4x + 12$ is factorised as $4(x+3)$.

Reason (R): $4$ is the highest common factor (HCF) of $4x$ and $12$.

---

To Determine:

Whether Assertion (A) and Reason (R) are true, and if Reason (R) is the correct explanation for Assertion (A).

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Solution:

Let's analyse the Assertion and the Reason separately.

Analysis of Assertion (A):

The expression is $4x + 12$. We need to factorise it by taking out common factors.

The terms are $4x$ and $12$.

Let's find the factors of each term:

Factors of $4x$: The numerical part 4 has factors 1, 2, 4. The variable part $x$ has factor $x$. So, factors are 1, 2, 4, $x$, $2x$, $4x$.

Factors of 12: The numerical part 12 has factors 1, 2, 3, 4, 6, 12.

The common numerical factors of 4 and 12 are 1, 2, and 4.

The common variable factor is 1 (since 12 does not have $x$).

The highest common factor (HCF) of the numerical coefficients 4 and 12 is 4.

The highest common factor of the variable parts is 1.

Thus, the HCF of $4x$ and 12 is $4 \times 1 = 4$.

To factorise $4x + 12$, we take out the HCF, 4:

$4x + 12 = 4(\frac{4x}{4}) + 4(\frac{12}{4})$

$= 4(x) + 4(3)$

$= 4(x + 3)$

The factorisation of $4x + 12$ is indeed $4(x+3)$.

Therefore, Assertion (A) is true.

Analysis of Reason (R):

The reason states that 4 is the highest common factor (HCF) of $4x$ and $12$.

From our analysis above, we found that the HCF of $4x$ and 12 is 4.

Therefore, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) claims a specific factorisation of $4x + 12$. Reason (R) identifies the HCF of the terms in the expression. The process of factorising by taking out the common factor involves finding the HCF and writing the expression as the HCF multiplied by the remaining terms. Since 4 is the HCF, taking out 4 as the common factor leads directly to the factorisation $4(x+3)$. Therefore, Reason (R) explains why Assertion (A) is true.

Both A and R are true, and R is the correct explanation of A.

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Answer:

Based on the analysis, both the assertion and the reason are true, and the reason correctly explains the assertion.

The correct option is (A).

Given:

Four algebraic expressions and four potential factorised forms.

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To Match:

Match each expression with its correct factorised form.

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Solution:

We will factorise each expression individually and then match it with the corresponding option from the list (a), (b), (c), (d).

(i) Expression: $x^2 - 1$

This is a difference of squares, $a^2 - b^2$, where $a=x$ and $b=1$.

Using the identity $a^2 - b^2 = (a+b)(a-b)$, we get:

$x^2 - 1 = (x+1)(x-1)$

This matches option (b) $(x+1)(x-1)$.

So, (i) $\to$ (b).

(ii) Expression: $x^2 + 2x + 1$

This is a trinomial of the form $ax^2 + bx + c$. We check if it is a perfect square trinomial $a^2 + 2ab + b^2 = (a+b)^2$.

The first term is $x^2$, which is $(x)^2$. So $a=x$.

The last term is 1, which is $(1)^2$. So $b=1$.

The middle term is $2x$. Let's check $2ab = 2(x)(1) = 2x$. This matches the middle term.

So, $x^2 + 2x + 1 = (x+1)^2$.

This matches option (a) $(x+1)^2$.

So, (ii) $\to$ (a).

(iii) Expression: $x^2 - 2x + 1$

This is a trinomial. We check if it is a perfect square trinomial $a^2 - 2ab + b^2 = (a-b)^2$.

The first term is $x^2$, which is $(x)^2$. So $a=x$.

The last term is 1, which is $(1)^2$. So $b=1$.

The middle term is $-2x$. Let's check $-2ab = -2(x)(1) = -2x$. This matches the middle term.

So, $x^2 - 2x + 1 = (x-1)^2$.

This matches option (d) $(x-1)^2$.

So, (iii) $\to$ (d).

(iv) Expression: $xy + y + x + 1$

This expression has four terms. We can try factorisation by grouping.

Group the first two terms and the last two terms:

$(xy + y) + (x + 1)$

Factor out the common factor from the first group ($xy + y$): The common factor is $y$.

$y(x + 1)$

Factor out the common factor from the second group ($x + 1$): The common factor is 1.

$1(x + 1)$

The expression becomes:

$y(x + 1) + 1(x + 1)$

Now, we have a common binomial factor $(x + 1)$. Factor it out:

$(x + 1)(y + 1)$

This matches option (c) $(x+1)(y+1)$.

So, (iv) $\to$ (c).

Based on our factorisations, the matches are:

(i) $\to$ (b)

(ii) $\to$ (a)

(iii) $\to$ (d)

(iv) $\to$ (c)

Now, let's check which of the given options (A), (B), (C), (D) corresponds to these matches.

Option (A): (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)

Option (B): (i)-(a), (ii)-(b), (iii)-(d), (iv)-(c)

Option (C): (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

Option (D): (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)

Comparing our matches with the options, we see that Option (A) correctly lists all the matches.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The area of the rectangular plot is given by the expression $x^2 + 5x + 6$ square meters.

---

To Find:

The type of polynomial that represents the area of the plot.

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Solution:

The area of the plot is given as the expression $x^2 + 5x + 6$.

A polynomial is classified based on the number of terms it contains:

- A polynomial with one term is called a monomial (e.g., $5x$, $7$).

- A polynomial with two terms is called a binomial (e.g., $x+y$, $3x^2 - 4$).

- A polynomial with three terms is called a trinomial (e.g., $ax^2 + bx + c$, $x^2 + 5x + 6$).

- A polynomial with four terms is called a quadrinomial (e.g., $a+b+c+d$).

The given expression for the area is $x^2 + 5x + 6$. This expression consists of three terms: $x^2$, $5x$, and $6$.

Since the expression has three terms, it is a trinomial.

Comparing this with the given options:

(A) Monomial

(B) Binomial

(C) Trinomial - This matches our classification.

(D) Quadrinomial

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

Area of the rectangular plot $= x^2 + 5x + 6$ square meters.

Length of the plot $= (x+3)$ meters.

Formula: Area = length $\times$ width.

---

To Find:

The expression for the width of the plot by dividing the area by the length.

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Solution:

We are given the area of the rectangle and its length. The relationship between area, length, and width is:

Area = Length $\times$ Width

To find the width, we can rearrange the formula:

Width $= \frac{\text{Area}}{\text{Length}}$

Substitute the given expressions for Area and Length:

Width $= \frac{x^2 + 5x + 6}{x+3}$

To perform this division, we can factorise the numerator, $x^2 + 5x + 6$.

We need to find two numbers that multiply to the constant term (6) and add up to the coefficient of the middle term (5).

Let the two numbers be $p$ and $q$. We need $p \times q = 6$ and $p + q = 5$.

The pairs of integers that multiply to 6 are (1, 6), (2, 3), (-1, -6), (-2, -3).

Let's check their sums:

$1 + 6 = 7$

$2 + 3 = 5$

$-1 + (-6) = -7$

$-2 + (-3) = -5$

The numbers 2 and 3 satisfy the conditions $2 \times 3 = 6$ and $2 + 3 = 5$.

So, we can factor the trinomial $x^2 + 5x + 6$ as $(x+2)(x+3)$.

Now, substitute the factored form back into the division expression:

Width $= \frac{(x+2)(x+3)}{x+3}$

Assuming $x+3 \neq 0$, we can cancel the common factor $(x+3)$ from the numerator and the denominator:

Width $= \frac{(x+2)\cancel{(x+3)}}{\cancel{(x+3)}}$

Width $= x+2$

The expression for the width of the plot is $(x+2)$ meters.

Compare this result with the given options:

(A) $(x+2)$ meters - This matches our result.

(B) $(x+3)$ meters

(C) $(x-2)$ meters

(D) $(x-3)$ meters

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

Area of the rectangular plot $= x^2 + 5x + 6$ square meters.

Length of the plot $= (x+3)$ meters.

Given value of $x = 10$ meters.

---

To Find:

The actual area of the plot when $x = 10$ meters.

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Solution:

We can find the actual area by substituting the given value of $x=10$ into the expression for the area.

Area $= x^2 + 5x + 6$

Substitute $x=10$:

Area $= (10)^2 + 5(10) + 6$

Area $= 100 + 50 + 6$

Area $= 156$ square meters.

This calculation is shown in Option (A).

Alternatively, we know that Area = Length $\times$ Width. From the case study, Length $= x+3$. From the previous question (Question 18), we found that Width $= x+2$.

Substitute $x=10$ into the expressions for Length and Width:

Length $= 10 + 3 = 13$ meters

Width $= 10 + 2 = 12$ meters

Now, calculate the area by multiplying the length and width:

Area $= \text{Length} \times \text{Width} = 13 \times 12$

Area $= 156$ square meters.

This calculation is shown in Option (B).

Both methods yield the same result, which is 156 square meters. Option (C) states this numerical result.

Since Options (A), (B), and (C) all correctly describe valid ways to calculate the area or state the correct resulting area, Option (D), which says "All of the above are correct ways to calculate," is the most appropriate answer.

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Answer:

The correct option is (D).

Given:

The sentence to complete: "Finding the factors of an algebraic expression is the reverse process of _________."

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To Determine:

The operation which is the reverse process of finding the factors of an algebraic expression.

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Solution:

Factorisation is the process of expressing an algebraic expression as a product of its factors.

Let's consider an example to understand the relationship between factorisation and other algebraic operations.

Suppose we have two algebraic expressions (factors), say $(x+2)$ and $(x+3)$. When we multiply these factors, we get:

$(x+2)(x+3) = x \cdot (x+3) + 2 \cdot (x+3)$

$= x^2 + 3x + 2x + 6$

$= x^2 + (3+2)x + 6$

$= x^2 + 5x + 6$

So, the product of the factors $(x+2)$ and $(x+3)$ is the expression $x^2 + 5x + 6$. This is the process of multiplication.

Factorisation is the process of starting with the expression $x^2 + 5x + 6$ and finding that it can be written as the product $(x+2)(x+3)$. In other words, we are reversing the multiplication process to find the original factors.

Let's look at the other options:

Addition and subtraction are operations that combine or separate terms, not typically factors in the sense meant here.

Division can be used as a tool to find factors (e.g., if you know one factor of a polynomial, you can divide the polynomial by that factor to find the other), but the process of factorisation itself is the reverse of the multiplication of algebraic expressions.

Therefore, finding the factors of an algebraic expression is the reverse process of multiplication.

Comparing this with the given options:

(A) Addition

(B) Subtraction

(C) Multiplication - This is the correct answer.

(D) Division

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

The two algebraic expressions are $14p^2q$ and $21pq^2$.

---

To Find:

The Highest Common Factor (HCF) of the given algebraic expressions.

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Solution:

To find the HCF of algebraic expressions, we find the HCF of the numerical coefficients and the HCF of the variable parts separately, and then multiply them.

First, find the HCF of the numerical coefficients, which are 14 and 21.

The prime factorization of 14 is $2 \times 7$.

The prime factorization of 21 is $3 \times 7$.

The common prime factor is 7. The highest power of the common prime factor is $7^1 = 7$.

So, the HCF of 14 and 21 is 7.

Next, find the HCF of the variable parts, which are $p^2q$ and $pq^2$.

For each variable, the HCF is the variable raised to the lowest power present in either term.

For the variable $p$: The powers are $p^2$ and $p^1$. The lowest power is $p^1$, which is $p$.

For the variable $q$: The powers are $q^1$ and $q^2$. The lowest power is $q^1$, which is $q$.

The HCF of the variable parts is $p^1q^1 = pq$.

The HCF of the entire expressions is the product of the HCF of the numerical part and the HCF of the variable part.

HCF$(14p^2q, 21pq^2) = (\text{HCF of } 14 \text{ and } 21) \times (\text{HCF of } p^2q \text{ and } pq^2)$

HCF $= 7 \times pq = 7pq$.

Thus, the HCF of $14p^2q$ and $21pq^2$ is $7pq$.

Comparing this with the given options:

(A) $7pq$ - This matches our result.

(B) $7p^2q^2$

(C) $42p^2q^2$

(D) $7pq^2$

Option (A) is the correct answer.

---

Answer:

The correct option is (A).

Given:

The algebraic expression is $5y(y-2) - 7(y-2)$.

---

To Find:

Factorise the given expression.

---

Solution:

The given expression is $5y(y-2) - 7(y-2)$.

This expression consists of two terms:

Term 1: $5y(y-2)$

Term 2: $-7(y-2)$

We can observe that the binomial $(y-2)$ is a common factor in both terms.

We can factor out the common binomial factor $(y-2)$ from the expression.

$5y(y-2) - 7(y-2) = (y-2) \times (5y - 7)$

Alternatively, we can write the factored form as $(5y - 7)(y - 2)$.

Let's verify this by expanding the factored form:

$(5y - 7)(y - 2) = 5y(y - 2) - 7(y - 2)$

$= 5y^2 - 10y - 7y + 14$

$= 5y^2 - 17y + 14$

Wait, my expansion does not match the original form $5y(y-2) - 7(y-2)$. The original form is already partially factored. The request is to factorise the entire expression. The expression is the subtraction of two terms, where each term contains the common factor $(y-2)$.

The structure of the expression is (Something) $\times (y-2) - (Something Else) \times (y-2)$.

The "Something" is $5y$ and the "Something Else" is $7$.

Factoring out the common term $(y-2)$ gives:

$(y-2) \times (5y \text{ from the first term } - 7 \text{ from the second term})$

So, $5y(y-2) - 7(y-2) = (y-2)(5y - 7)$.

This is the standard way to factor when a binomial is common.

Comparing this with the given options:

(A) $(5y-7)(y-2)$ - This matches our result.

(B) $(5y+7)(y-2)$

(C) $(5y-7)(y+2)$

(D) $(5y+7)(y+2)$

Option (A) is the correct answer.

---

Answer:

The correct option is (A).

Given:

The expression is $(a+b)^2 - (a-b)^2$.

---

To Find:

Simplify the given expression and find its factors, then select the correct option based on the results.

---

Solution:

First, let's simplify the expression $(a+b)^2 - (a-b)^2$.

We can use the algebraic identities for perfect squares:

$(a+b)^2 = a^2 + 2ab + b^2$

$(a-b)^2 = a^2 - 2ab + b^2$

Now, substitute these expansions into the given expression:

$(a+b)^2 - (a-b)^2 = (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)$

Distribute the negative sign to the terms inside the second bracket:

$= a^2 + 2ab + b^2 - a^2 + 2ab - b^2$

Group the like terms:

$= (a^2 - a^2) + (2ab + 2ab) + (b^2 - b^2)$

Combine the like terms:

$= 0 + 4ab + 0$

$= 4ab$

Alternatively, we can use the difference of squares identity: $X^2 - Y^2 = (X - Y)(X + Y)$.

Let $X = (a+b)$ and $Y = (a-b)$.

Then $(a+b)^2 - (a-b)^2 = [(a+b) - (a-b)][(a+b) + (a-b)]$

Simplify the terms within the square brackets:

$[(a+b) - (a-b)] = a+b - a + b = 2b$

$[(a+b) + (a-b)] = a+b + a - b = 2a$

Multiply the simplified terms:

$(2b)(2a) = 4ab$

The simplified expression is $4ab$.

---

Now, let's consider the factors of the simplified expression, $4ab$.

The factors of $4ab$ are the terms that divide $4ab$ evenly. These include the numerical factors of 4 (which are 1, 2, 4) and the variable factors ($a$, $b$) and all possible products of these numerical and variable factors.

The factors are $1, 2, 4, a, b, ab, 2a, 2b, 4a, 4b, 2ab, 4ab$.

Let's evaluate the given options in relation to the simplified expression ($4ab$) and its factors:

(A) $4ab$ - This is the simplified form of the expression. The expression itself is also a factor.

(B) $(2a)(2b)$ - This is a product of two factors, $2a$ and $2b$, which equals $4ab$. $2a$ is a factor of $4ab$, and $2b$ is a factor of $4ab$. The product is a factored form of the expression.

(C) $4ab$, factors are $4, a, b$ and their combinations - This correctly states the simplified expression ($4ab$) and describes its factors. $4, a, b$ are indeed factors, and "their combinations" (like $ab, 4a, 4b, 4ab$) are also factors.

All three options (A), (B), and (C) provide correct information related to the simplification and/or factorisation of the expression $(a+b)^2 - (a-b)^2$. Option (D) claims that all of the above are related.

Since A is the simplified result, B is a valid factored form equal to the simplified result, and C correctly states the simplified result and describes its factors, all three statements are true and related to the problem.

Therefore, the most comprehensive correct answer is that all of the above are related to the simplification and factors.

Option (D) is the correct answer.

---

Answer:

The correct option is (D).

Given:

The expression to be divided is $(25a^2 - 4b^2)$ by $(5a - 2b)$.

---

To Find:

The result of the division $(25a^2 - 4b^2) \div (5a - 2b)$.

---

Solution:

We need to compute the division $\frac{25a^2 - 4b^2}{5a - 2b}$.

Let's examine the numerator, $25a^2 - 4b^2$. We can rewrite each term as a square:

$25a^2 = (5a)^2$

$4b^2 = (2b)^2$

So, the numerator is a difference of squares: $(5a)^2 - (2b)^2$.

Using the algebraic identity for the difference of squares, $X^2 - Y^2 = (X - Y)(X + Y)$, with $X = 5a$ and $Y = 2b$, we can factor the numerator:

$(5a)^2 - (2b)^2 = (5a - 2b)(5a + 2b)$

Now, substitute this factored form of the numerator into the division expression:

$\frac{25a^2 - 4b^2}{5a - 2b} = \frac{(5a - 2b)(5a + 2b)}{5a - 2b}$

Assuming that the denominator $(5a - 2b)$ is not equal to zero, we can cancel out the common factor $(5a - 2b)$ from the numerator and the denominator:

$\frac{\cancel{(5a - 2b)}(5a + 2b)}{\cancel{(5a - 2b)}} = 5a + 2b$

Thus, the result of the division is $5a + 2b$.

Compare this result with the given options:

(A) $5a - 2b$

(B) $5a + 2b$ - This matches our result.

(C) $25a + 4b$

(D) $25a - 4b$

Option (B) is the correct answer.

---

Answer:

The correct option is (B).

Given:

The product of two algebraic expressions is $18x^2 - 9x$.

One of the expressions is $9x$.

---

To Find:

The other expression.

---

Solution:

Let the two algebraic expressions be $A$ and $B$.

We are given that the product of the two expressions is $18x^2 - 9x$. So, $A \times B = 18x^2 - 9x$.

We are also given that one of the expressions is $9x$. Let's say $A = 9x$.

Then, we have $9x \times B = 18x^2 - 9x$.

To find the other expression, $B$, we need to divide the product by the known expression:

$B = \frac{18x^2 - 9x}{9x}$

To perform this division, we divide each term in the numerator by the denominator.

$B = \frac{18x^2}{9x} - \frac{9x}{9x}$

Divide the first term:

$\frac{18x^2}{9x} = \frac{18}{9} \times \frac{x^2}{x} = 2 \times x^{2-1} = 2x^1 = 2x$

Divide the second term:

$\frac{9x}{9x} = \frac{9}{9} \times \frac{x}{x} = 1 \times x^{1-1} = 1 \times x^0 = 1 \times 1 = 1$ (assuming $x \neq 0$)

Substitute these results back into the expression for $B$:

$B = 2x - 1$

So, the other expression is $2x - 1$.

We can verify this by multiplying the two expressions: $9x \times (2x - 1) = 9x \times 2x - 9x \times 1 = 18x^2 - 9x$, which is the given product.

Comparing our result with the given options:

(A) $2x^2 - x$

(B) $2x - 1$ - This matches our result.

(C) $18x^2$

(D) $9x$

Option (B) is the correct answer.

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Answer:

The correct option is (B).

Given:

The algebraic expression is $p^4 - 81$.

Four potential factors are given in the options.

---

To Find:

Which of the given options is NOT a factor of $p^4 - 81$.

---

Solution:

To determine which expression is not a factor of $p^4 - 81$, we first need to factorise $p^4 - 81$.

The expression $p^4 - 81$ is a difference of squares, as $p^4 = (p^2)^2$ and $81 = 9^2$.

Using the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$, where $a = p^2$ and $b = 9$, we have:

$p^4 - 81 = (p^2)^2 - 9^2 = (p^2 - 9)(p^2 + 9)$

Now, observe the factor $(p^2 - 9)$. This is also a difference of squares, as $p^2 = (p)^2$ and $9 = 3^2$.

Using the difference of squares identity again, where $a = p$ and $b = 3$, we have:

$p^2 - 9 = p^2 - 3^2 = (p - 3)(p + 3)$

Substitute this factorisation back into the expression for $p^4 - 81$:

$p^4 - 81 = (p^2 - 9)(p^2 + 9) = (p - 3)(p + 3)(p^2 + 9)$

The complete factorisation of $p^4 - 81$ into irreducible factors over real numbers is $(p - 3)(p + 3)(p^2 + 9)$.

The factors of $p^4 - 81$ include 1, $(p-3)$, $(p+3)$, $(p^2+9)$, and any product of these factors.

Let's check the given options:

(A) $p^2 + 9$: This is one of the factors in the complete factorisation. So, it IS a factor.

(B) $p + 3$: This is one of the factors in the complete factorisation. So, it IS a factor.

(C) $p - 3$: This is one of the factors in the complete factorisation. So, it IS a factor.

(D) $p^2 - 9p + 81$: This expression does not appear in our factorisation, nor can it be formed by multiplying the prime factors $(p - 3)$, $(p + 3)$, and $(p^2 + 9)$. For instance, if we were to multiply $(p-3)(p+3) = p^2 - 9$, and we multiply by $(p^2+9)$, we get $p^4-81$. There is no direct product that yields $p^2 - 9p + 81$. To confirm it's not a factor, one could perform polynomial division or check the roots, but based on the standard factorisation process, it is clear that this expression is not one of the simple factors or their combination resulting in the original polynomial.

Therefore, $p^2 - 9p + 81$ is NOT a factor of $p^4 - 81$.

Option (D) is the correct answer.

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Answer:

The correct option is (D).

Given:

The trinomial is $x^2 - 13x + 30$.

A list of potential factors in options (A) to (E).

---

To Find:

Which of the given options are factors of the trinomial $x^2 - 13x + 30$.

---

Solution:

To find the factors of the trinomial $x^2 - 13x + 30$, which is in the form $x^2 + bx + c$, we need to find two numbers, let's call them $p$ and $q$, such that their product is equal to the constant term ($c$) and their sum is equal to the coefficient of the $x$ term ($b$).

In the expression $x^2 - 13x + 30$, we have $b = -13$ and $c = 30$.

We are looking for two numbers $p$ and $q$ such that:

$p \times q = 30$

$p + q = -13$

Since the product ($p \times q = 30$) is positive and the sum ($p+q = -13$) is negative, both numbers $p$ and $q$ must be negative.

Let's list pairs of negative integers whose product is 30 and check their sums:

Pairs that multiply to 30: (-1, -30), (-2, -15), (-3, -10), (-5, -6)

Calculate the sum for each pair:

$-1 + (-30) = -31$

$-2 + (-15) = -17$

$-3 + (-10) = -13$

$-5 + (-6) = -11$

The pair of numbers that adds up to -13 and multiplies to 30 is -3 and -10.

So, we can factor the trinomial as $(x+p)(x+q)$, which is $(x+(-3))(x+(-10)) = (x-3)(x-10)$.

The factors of $x^2 - 13x + 30$ are $(x-3)$ and $(x-10)$.

Now, let's compare these factors with the given options:

(A) $(x-3)$: This matches one of the factors we found.

(B) $(x-10)$: This matches the other factor we found.

(C) $(x+3)$: This is not a factor.

(D) $(x+10)$: This is not a factor.

(E) $(x-13)$: This is not a factor.

The question asks to select all options that are factors.

Therefore, options (A) and (B) are the correct choices.

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Answer:

The factors of $x^2 - 13x + 30$ are $(x-3)$ and $(x-10)$.

The correct options are (A) and (B).

Given:

The volume of the rectangular water tank is given by the expression $10(x^2 - 4)$ cubic meters.

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To Find:

The factorised form of the volume expression $10(x^2 - 4)$.

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Solution:

The given expression for the volume is $10(x^2 - 4)$.

To factorise this expression, we need to factorise the term inside the parenthesis, which is $x^2 - 4$.

The expression $x^2 - 4$ is a difference of two perfect squares, as $x^2 = (x)^2$ and $4 = 2^2$.

We can use the algebraic identity for the difference of squares: $a^2 - b^2 = (a - b)(a + b)$.

In the expression $x^2 - 4$, we have $a = x$ and $b = 2$.

Applying the identity to $x^2 - 4$:

$x^2 - 4 = (x)^2 - (2)^2 = (x - 2)(x + 2)$

Now, substitute this factored form of $(x^2 - 4)$ back into the original volume expression $10(x^2 - 4)$:

$10(x^2 - 4) = 10 \times (x - 2)(x + 2)$

The factorised form of the volume expression is $10(x+2)(x-2)$ (the order of factors does not matter).

Compare this result with the given options:

(A) $10(x-2)^2 = 10(x^2 - 4x + 4)$

(B) $10(x+2)^2 = 10(x^2 + 4x + 4)$

(C) $10(x+2)(x-2) = 10(x^2 - 4)$ - This matches our factorised form.

(D) $10x(x-4/x) = 10x^2 - 40$

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

Volume of the rectangular tank $= 10(x^2 - 4)$ cubic meters.

Height of the tank $= 10$ meters.

Length of the tank $= (x+2)$ meters.

Formula for volume of a rectangular tank = length $\times$ width $\times$ height.

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To Find:

The expression for the width of the tank.

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Solution:

The volume of a rectangular tank is given by the formula:

Volume = Length $\times$ Width $\times$ Height

We are given the volume, length, and height, and we need to find the width. We can rearrange the formula to solve for the width:

Width $= \frac{\text{Volume}}{\text{Length} \times \text{Height}}$

Substitute the given expressions into this formula:

Width $= \frac{10(x^2 - 4)}{(x+2) \times 10}$

We can simplify the expression by cancelling the common factor of 10 in the numerator and the denominator:

Width $= \frac{\cancel{10}(x^2 - 4)}{(x+2) \times \cancel{10}}$

Width $= \frac{x^2 - 4}{x+2}$

Now, we need to simplify the fraction $\frac{x^2 - 4}{x+2}$. The numerator $x^2 - 4$ is a difference of squares, which can be factored using the identity $a^2 - b^2 = (a-b)(a+b)$.

$x^2 - 4 = x^2 - 2^2 = (x - 2)(x + 2)$

Substitute this factored form back into the expression for the width:

Width $= \frac{(x - 2)(x + 2)}{x + 2}$

Assuming $(x+2) \neq 0$, we can cancel the common factor $(x+2)$ from the numerator and the denominator:

Width $= \frac{(x - 2)\cancel{(x + 2)}}{\cancel{(x + 2)}}$

Width $= x - 2$

The expression for the width of the tank is $(x-2)$ meters.

Comparing our result with the given options:

(A) $\frac{10(x+2)(x-2)}{10(x+2)} = (x-2)$ meters - This option shows the correct calculation leading to the result $(x-2)$ meters.

(B) $(x+2)$ meters

(C) $10(x-2)$ meters

(D) $(x-4)$ meters

Option (A) presents the division step which results in the correct expression for the width.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

Volume of the rectangular tank $= 10(x^2 - 4)$ cubic meters.

Height of the tank $= 10$ meters.

Length of the tank $= (x+2)$ meters.

Value of $x = 5$ meters.

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To Find:

The actual dimensions (length, width, height) of the tank when $x=5$ meters.

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Solution:

We are given the expression for the length and the value of the height. From the previous question (Question 29), we found the expression for the width by dividing the volume by the product of length and height.

Length $= x+2$ meters.

Height $= 10$ meters.

Width $= x-2$ meters (from Question 29).

Now, we substitute the given value of $x=5$ meters into the expressions for length and width.

Calculate the actual length:

Length $= x+2 = 5 + 2 = 7$ meters.

Calculate the actual width:

Width $= x-2 = 5 - 2 = 3$ meters.

The height is given directly:

Height $= 10$ meters.

So, the actual dimensions of the tank when $x=5$ meters are Length = 7 m, Width = 3 m, and Height = 10 m.

Let's check if these dimensions give the correct volume for $x=5$.

Volume formula in terms of dimensions: Length $\times$ Width $\times$ Height $= 7 \times 3 \times 10 = 21 \times 10 = 210$ cubic meters.

Volume expression in terms of $x$: $10(x^2 - 4)$. Substitute $x=5$:

Volume $= 10((5)^2 - 4) = 10(25 - 4) = 10(21) = 210$ cubic meters.

The volume calculated using the dimensions matches the volume calculated using the expression with $x=5$. This confirms our dimensions are correct.

Compare the calculated dimensions with the given options:

(A) Length = 7 m, Width = 3 m, Height = 10 m - This matches our calculated dimensions.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The expression to be divided is $(a^2 + ab)$ by $a$. This can be written as $\frac{a^2 + ab}{a}$.

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To Find:

The result of the division $\frac{a^2 + ab}{a}$.

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Solution:

To divide a polynomial by a monomial, we can divide each term of the polynomial by the monomial divisor.

The expression is $\frac{a^2 + ab}{a}$.

We can rewrite this as the sum of individual fractions:

$\frac{a^2}{a} + \frac{ab}{a}$

Now, perform each division separately.

First term: $\frac{a^2}{a}$

Using the rule of exponents $\frac{x^m}{x^n} = x^{m-n}$, we have:

$\frac{a^2}{a^1} = a^{2-1} = a^1 = a$

Second term: $\frac{ab}{a}$

Assuming $a \neq 0$, we can cancel the common factor $a$:

$\frac{\cancel{a}b}{\cancel{a}} = b$

Combine the results of the individual divisions:

$a + b$

Thus, $(a^2 + ab) \div a = a + b$.

Comparing this result with the given options:

(A) $a+ab$

(B) $a+b$ - This matches our result.

(C) $a^2+b$

(D) $b$

Option (B) is the correct answer.

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Answer:

The correct option is (B).

Given:

The algebraic expression is $15pq + 15 + 9q + 25p$.

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To Find:

Factorise the given expression by grouping the terms.

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Solution:

We need to factorise the expression $15pq + 15 + 9q + 25p$ by grouping. We can rearrange and group the terms in different ways to find common factors.

Let's rearrange the terms to group those with common factors:

Option 1: Group terms with $p$ and terms without $p$ (or other combinations). This doesn't seem straightforward due to the constant term.

Option 2: Rearrange to group terms that share common variables or numbers. Let's try grouping $15pq$ with $9q$ and $25p$ with $15$.

Rearrange the expression: $15pq + 9q + 25p + 15$

Group the terms: $(15pq + 9q) + (25p + 15)$

Factor out the common factor from the first group $(15pq + 9q)$: The common numerical factor of 15 and 9 is 3. The common variable factor is $q$. The common factor is $3q$.

$15pq + 9q = 3q(5p) + 3q(3) = 3q(5p + 3)$

Factor out the common factor from the second group $(25p + 15)$: The common numerical factor of 25 and 15 is 5.

$25p + 15 = 5(5p) + 5(3) = 5(5p + 3)$

Substitute these factored forms back into the grouped expression:

$3q(5p + 3) + 5(5p + 3)$

Now, we see that $(5p + 3)$ is a common binomial factor in both terms. Factor out $(5p + 3)$:

$(5p + 3)(3q + 5)$

This is the factorised form of the expression.

Alternatively, let's try another grouping: Group $15pq$ with $25p$ and $9q$ with $15$.

Rearrange the expression: $15pq + 25p + 9q + 15$

Group the terms: $(15pq + 25p) + (9q + 15)$

Factor out the common factor from the first group $(15pq + 25p)$: The common numerical factor of 15 and 25 is 5. The common variable factor is $p$. The common factor is $5p$.

$15pq + 25p = 5p(3q) + 5p(5) = 5p(3q + 5)$

Factor out the common factor from the second group $(9q + 15)$: The common numerical factor of 9 and 15 is 3.

$9q + 15 = 3(3q) + 3(5) = 3(3q + 5)$

Substitute these factored forms back into the grouped expression:

$5p(3q + 5) + 3(3q + 5)$

Now, we see that $(3q + 5)$ is a common binomial factor in both terms. Factor out $(3q + 5)$:

$(3q + 5)(5p + 3)$

Both grouping methods lead to the same factorisation: $(5p + 3)(3q + 5)$.

Comparing our result with the given options:

(A) $(5p+3)(3q+5)$ - This matches our factorised form.

(B) $(5p+5)(3q+3)$

(C) $(15p+9)(q+?) ...$ cannot be factorised this way

(D) $(5p+3q)(3+5)$

Option (A) is the correct answer.

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Answer:

The correct option is (A).

Given:

The expression is $\frac{x^2 - 25}{x - 5}$.

The value of $x$ is 6.

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To Evaluate:

Evaluate the given expression when $x = 6$.

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Solution:

We need to find the value of the expression $\frac{x^2 - 25}{x - 5}$ when $x = 6$.

There are two common ways to evaluate this expression at a specific value of $x$ (provided the denominator is not zero):

Method 1: Direct Substitution

Substitute $x = 6$ directly into the given expression:

$\frac{x^2 - 25}{x - 5} = \frac{(6)^2 - 25}{6 - 5}$

Calculate the numerator and the denominator:

Numerator: $6^2 - 25 = 36 - 25 = 11$

Denominator: $6 - 5 = 1$

So, the expression evaluates to:

$\frac{11}{1} = 11$

This method is shown in Option (A).

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Method 2: Factorise and Simplify First, then Substitute

Factorise the numerator of the expression, $x^2 - 25$. This is a difference of squares, $x^2 - 5^2$.

Using the identity $a^2 - b^2 = (a - b)(a + b)$, where $a = x$ and $b = 5$, we get:

$x^2 - 25 = (x - 5)(x + 5)$

Now, rewrite the expression with the factored numerator:

$\frac{x^2 - 25}{x - 5} = \frac{(x - 5)(x + 5)}{x - 5}$

Assuming $x - 5 \neq 0$, we can cancel the common factor $(x - 5)$ from the numerator and the denominator. When $x = 6$, $x - 5 = 6 - 5 = 1 \neq 0$, so the cancellation is valid.

$\frac{\cancel{(x - 5)}(x + 5)}{\cancel{(x - 5)}} = x + 5$

Now, substitute $x = 6$ into the simplified expression $x + 5$:

$x + 5 = 6 + 5 = 11$

This method is shown in Option (B).

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Both methods yield the same result, 11. Option (C) states the final numerical value.

Since Option (A) correctly shows the direct substitution and calculation, Option (B) correctly shows the factorisation, simplification, and substitution, and Option (C) correctly states the final result obtained by these methods, all three options are related to the correct evaluation of the expression.

Therefore, Option (D) is the most appropriate answer, as it states that all of the above give the correct value or show a correct way to calculate it.

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Answer:

The correct option is (D).

Given:

$36x^2y^3 - 48x^3y^2$

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Solution:

Find the HCF of $36x^2y^3$ and $48x^3y^2$.

HCF of 36 and 48 is 12.

HCF of $x^2y^3$ and $x^3y^2$ is $x^2y^2$ (taking the lowest power of each variable).

The HCF of the expression is $12x^2y^2$.

Factorise the expression by taking out the HCF:

$36x^2y^3 - 48x^3y^2 = 12x^2y^2 \left( \frac{36x^2y^3}{12x^2y^2} - \frac{48x^3y^2}{12x^2y^2} \right)$

$= 12x^2y^2 (3y - 4x)$

The expression inside the bracket is $(3y - 4x)$.

We need the coefficient of the $xy$ term inside the bracket. The terms inside the bracket are $3y$ and $-4x$. There is no $xy$ term inside the bracket.

However, looking at the options (3, -3, 4, -4), it is likely the question intended to ask for the coefficient of either the $x$ term or the $y$ term.

The coefficient of the $y$ term inside the bracket is 3.

The coefficient of the $x$ term inside the bracket is -4.

Option (D) is -4.

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Answer:

Based on the likely intent of the question, the coefficient of the $x$ term inside the bracket is -4.

The correct option is (D).

Given:

Assertion (A): Division is the inverse operation of multiplication for algebraic expressions.

Reason (R): If $A \times B = C$, then $\frac{C}{B} = A$ (provided $B \neq 0$).

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To Determine:

Whether Assertion (A) and Reason (R) are true, and if Reason (R) is the correct explanation for Assertion (A).

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Solution:

Let's analyse the Assertion (A).

In mathematics, an inverse operation undoes the effect of another operation. For numbers, division is the inverse of multiplication because if you multiply a number by another (non-zero) number, you can divide the result by the second number to get back the first number. This principle extends to algebraic expressions. For example, if we multiply $(x+1)$ by $(x-1)$, we get $x^2 - 1$. If we then divide $x^2 - 1$ by $(x-1)$, we get $(x+1)$. Thus, division is indeed the inverse operation of multiplication for algebraic expressions (assuming the divisor is non-zero).

Therefore, Assertion (A) is true.

Now, let's analyse the Reason (R).

The statement "If $A \times B = C$, then $\frac{C}{B} = A$ (provided $B \neq 0$)" is the definition of division. It formally states that dividing the product $C$ by one factor $B$ gives the other factor $A$. This is precisely the property that makes division the inverse of multiplication.

Therefore, Reason (R) is true.

Finally, let's consider the relationship between Assertion (A) and Reason (R).

Reason (R) provides the fundamental definition or property that underpins the statement in Assertion (A). The fact that $\frac{C}{B} = A$ when $A \times B = C$ (for $B \neq 0$) is the core reason why division is considered the inverse operation of multiplication. Reason (R) explains the mechanism by which division undoes multiplication.

Therefore, Reason (R) is the correct explanation of Assertion (A).

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Answer:

Based on the analysis, both the assertion and the reason are true, and the reason correctly explains the assertion.

The correct option is (A).

Given:

The trinomial is $a^2 + 10a + 21$.

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To Find:

Which of the given options is NOT a factor of $a^2 + 10a + 21$.

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Solution:

To find the factors of the trinomial $a^2 + 10a + 21$, which is of the form $x^2 + bx + c$, we need to find two numbers that multiply to the constant term ($c=21$) and add up to the coefficient of the middle term ($b=10$).

Let the two numbers be $p$ and $q$. We need:

$p \times q = 21$

$p + q = 10$

We look for pairs of integers whose product is 21:

(1, 21), (3, 7), (-1, -21), (-3, -7).

Let's find the sum for each pair:

$1 + 21 = 22$

$3 + 7 = 10$

$-1 + (-21) = -22$

$-3 + (-7) = -10$

The pair of numbers that sums to 10 and multiplies to 21 is 3 and 7.

Therefore, the trinomial can be factored as:

$a^2 + 10a + 21 = (a + 3)(a + 7)$

The factors of $a^2 + 10a + 21$ are 1, $(a+3)$, $(a+7)$, and $(a+3)(a+7)$.

Now, let's compare these factors with the given options:

(A) $(a+3)$: This is a factor.

(B) $(a+7)$: This is a factor.

(C) $(a-3)$: This is NOT a factor.

(D) $(a+3)(a+7)$: This is the original expression in factored form, and the expression itself is considered a factor.

The option that is not a factor of $a^2 + 10a + 21$ is $(a-3)$.

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

The expression to be divided is $(x^2 + 6x + 9)$ by $(x+3)$.

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To Find:

The remainder when $(x^2 + 6x + 9)$ is divided by $(x+3)$.

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Solution:

We are asked to divide the polynomial $x^2 + 6x + 9$ by the polynomial $x+3$.

We can perform this division by factoring the numerator, if possible.

Consider the numerator $x^2 + 6x + 9$. This is a trinomial in the form $ax^2 + bx + c$. We can check if it is a perfect square trinomial of the form $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$.

The first term $x^2$ is the square of $x$, so we can consider $a=x$.

The last term 9 is the square of 3 ($3^2 = 9$), so we can consider $b=3$.

Let's check if the middle term $6x$ matches $2ab$ or $-2ab$ with $a=x$ and $b=3$.

$2ab = 2(x)(3) = 6x$.

The middle term matches $2ab$ and the other terms are $a^2$ and $b^2$. Thus, the trinomial $x^2 + 6x + 9$ is a perfect square trinomial and can be factored using the identity $a^2 + 2ab + b^2 = (a+b)^2$.

So, $x^2 + 6x + 9 = (x+3)^2$.

Now, we can write the division as:

$\frac{x^2 + 6x + 9}{x+3} = \frac{(x+3)^2}{x+3}$

Assuming the denominator $x+3$ is not equal to zero, we can simplify the expression by cancelling out one factor of $(x+3)$ from the numerator and the denominator.

$\frac{(x+3)\cancel{(x+3)}}{\cancel{(x+3)}} = x+3$

The result of the division is $x+3$ with no remainder. When a polynomial $P(x)$ is divided by $D(x)$ and the result is exactly $Q(x)$ (i.e., $P(x) = D(x) \times Q(x)$), the remainder is 0.

In this case, $(x^2 + 6x + 9) = (x+3) \times (x+3)$, so the remainder is 0.

Alternatively, we can use the Remainder Theorem. The Remainder Theorem states that when a polynomial $P(x)$ is divided by $(x-a)$, the remainder is $P(a)$.

Here, $P(x) = x^2 + 6x + 9$ and the divisor is $(x+3)$. This is in the form $(x-a)$ where $a = -3$.

According to the Remainder Theorem, the remainder is $P(-3)$.

$P(-3) = (-3)^2 + 6(-3) + 9$

$P(-3) = 9 - 18 + 9$

$P(-3) = 18 - 18$

$P(-3) = 0$

Both methods confirm that the remainder is 0.

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Answer:

The remainder when $(x^2 + 6x + 9)$ is divided by $(x+3)$ is 0.

The correct option is (C).

Given:

The incomplete sentence: "A factor that cannot be expressed as a product of factors other than itself and 1 is called an _________ factor."

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To Complete:

Complete the given sentence with the appropriate term.

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Solution:

The definition provided in the sentence describes a factor that cannot be broken down into simpler factors, except for trivial factors like itself and 1 (or -1 in some contexts, or constants in algebraic expressions).

In the context of numbers, such a factor is called a prime factor.

In the context of algebraic expressions (polynomials), such a factor is usually called an irreducible factor.

The options provided are:

(A) Common

(B) Prime or Irreducible

(C) Algebraic

(D) Monomial

Option (A) "Common" refers to a factor shared by two or more expressions.

Option (C) "Algebraic" refers to a factor that is an algebraic expression.

Option (D) "Monomial" refers to a factor that is a monomial (single term).

Option (B) "Prime or Irreducible" directly corresponds to the definition given in the sentence, covering both numerical and algebraic contexts where factors cannot be further decomposed non-trivially.

Therefore, the correct term to complete the sentence is "Prime or Irreducible".

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Answer:

The complete sentence is: A factor that cannot be expressed as a product of factors other than itself and 1 is called a Prime or Irreducible factor.

The correct option is (B).

Given:

Area of the trapezium $= 10x + 15y$ square units.

Height of the trapezium $= h = 5$ units.

One parallel side $= a = 2x$ units.

Let the other parallel side be $b$ units.

The formula for the area of a trapezium is Area $= \frac{1}{2}h(a+b)$.

The given equation based on this information is:

Area $= \frac{1}{2} \times 5 \times (2x + b) = 10x + 15y$

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To Find:

The expression for the other parallel side, $b$.

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Solution:

We are given the equation:

$\frac{1}{2} \times 5 \times (2x + b) = 10x + 15y$

Simplify the left side:

$\frac{5}{2}(2x + b) = 10x + 15y$

To isolate the term $(2x + b)$, multiply both sides of the equation by the reciprocal of $\frac{5}{2}$, which is $\frac{2}{5}$.

$\frac{2}{5} \times \frac{5}{2}(2x + b) = \frac{2}{5} \times (10x + 15y)$

$1 \times (2x + b) = \frac{2}{5}(10x) + \frac{2}{5}(15y)$

$2x + b = \frac{20x}{5} + \frac{30y}{5}$

Perform the divisions on the right side:

$2x + b = 4x + 6y$

Now, to find the expression for $b$, subtract $2x$ from both sides of the equation:

$b = 4x + 6y - 2x$

Combine the like terms ($4x$ and $-2x$):

$b = (4x - 2x) + 6y$

$b = 2x + 6y$

The expression for the other parallel side is $2x + 6y$ units.

Comparing this result with the given options:

(A) $2x + 3y$

(B) $4x + 6y$

(C) $2x + 6y$ - This matches our result.

(D) $4x + 3y$

Option (C) is the correct answer.

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Answer:

The correct option is (C).

Given:

The expression for the area of a trapezium is $ax + ay$.

---

To Find:

The common factor that can be taken out from the expression $ax + ay$.

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Solution:

To find the common factor in the expression $ax + ay$, we look for factors that are present in both terms.

The first term is $ax$. Its factors are $1, a, x, ax$.

The second term is $ay$. Its factors are $1, a, y, ay$.

The factors common to both $ax$ and $ay$ are $1$ and $a$.

The highest common factor (HCF) among these common factors is $a$.

To factorise the expression by taking out the common factor $a$, we divide each term by $a$ and write the result in the form $a(\dots)$.

$ax + ay = a \left( \frac{ax}{a} + \frac{ay}{a} \right)$

$= a(x + y)$

The common factor that is taken out is $a$.

Comparing this with the given options:

(A) $a$ - This is the common factor we found.

(B) $x+y$ - This is the binomial factor remaining after taking out the common factor $a$. It is a factor of the expression, but not the common factor taken out from both terms.

(C) $ay$ - This is a factor of the second term, but not generally a factor of the first term $ax$ unless $x=y$ (and $a \neq 0$).

(D) $x$ - This is a factor of the first term, but not generally a factor of the second term $ay$ unless $a=0$ or $y=x$.

The common factor that can be taken out from both terms $ax$ and $ay$ is $a$.

Option (A) is the correct answer.

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Answer:

The correct option is (A).

The given algebraic term is $12xy^2$.

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To find the factors of the algebraic term, we first find the prime factors of the numerical coefficient and then consider the variable factors.

The numerical coefficient is 12. Let's find its prime factors:

$\begin{array}{c|cc} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 12 is $2 \times 2 \times 3$.

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Now, let's write the given term as a product of its prime numerical factors and variable factors.

$12xy^2 = 2 \times 2 \times 3 \times x \times y \times y$

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The factors of the algebraic term are the individual quantities that are multiplied together to form the term.

Therefore, the factors of $12xy^2$ are $2, 2, 3, x, y,$ and $y$.

Given: The algebraic terms are $6a^2b$ and $9ab^2$.

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To Find: The greatest common factor (GCF) of $6a^2b$ and $9ab^2$.

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Solution:

First, let's find the prime factors of each term.

For the term $6a^2b$:

The numerical coefficient is 6. Its prime factorization is:

$\begin{array}{c|cc} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $6 = 2 \times 3$.

The variable part is $a^2b = a \times a \times b$.

Thus, the factors of $6a^2b$ are $2, 3, a, a, b$.

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For the term $9ab^2$:

The numerical coefficient is 9. Its prime factorization is:

$\begin{array}{c|cc} 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $9 = 3 \times 3$.

The variable part is $ab^2 = a \times b \times b$.

Thus, the factors of $9ab^2$ are $3, 3, a, b, b$.

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Now, let's list the factors of each term:

Factors of $6a^2b$: $2, 3, a, a, b$

Factors of $9ab^2$: $3, 3, a, b, b$

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The common factors are the factors that appear in the factorization of both terms.

Common numerical factor: $3$

Common variable factors: $a, b$

---

The greatest common factor (GCF) is the product of all the common factors.

GCF $= 3 \times a \times b = 3ab$.

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The greatest common factor of $6a^2b$ and $9ab^2$ is $3ab$.

Given: The algebraic expression is $5x + 10$.

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To Factorise: The expression $5x + 10$ by taking out the common factors.

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Solution:

The given expression has two terms: $5x$ and $10$.

Let's find the factors of each term separately.

Factors of the first term $5x$: $5 \times x$

Factors of the second term $10$: $2 \times 5$

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Now, let's identify the common factors in both terms.

The factors of $5x$ are $5$ and $x$.

The factors of $10$ are $2$ and $5$.

The factor that is common to both terms is $5$.

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We can rewrite the expression by showing the common factor:

$5x + 10 = (5 \times x) + (5 \times 2)$

Now, we take the common factor $5$ outside the parentheses.

$5x + 10 = 5(x + 2)$

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Thus, the factorised form of $5x + 10$ is $5(x + 2)$.

Given: The algebraic expression is $14a^2 - 21ab$.

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To Factorise: The expression $14a^2 - 21ab$ by taking out the common factors.

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Solution:

The given expression has two terms: $14a^2$ and $21ab$.

Let's find the prime factors of the numerical coefficients:

For 14:

$\begin{array}{c|cc} 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $14 = 2 \times 7$.

For 21:

$\begin{array}{c|cc} 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $21 = 3 \times 7$.

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Now, let's write the factors of each term:

Factors of the first term $14a^2$: $2 \times 7 \times a \times a$

Factors of the second term $21ab$: $3 \times 7 \times a \times b$

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Let's identify the common factors in both terms.

The common numerical factor is $7$.

The common variable factor is $a$.

The greatest common factor (GCF) of $14a^2$ and $21ab$ is the product of the common factors, which is $7 \times a = 7a$.

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Now, we rewrite the expression by taking out the GCF $7a$ from each term:

$14a^2 - 21ab = (7a \times 2a) - (7a \times 3b)$

Factor out the common factor $7a$:

$14a^2 - 21ab = 7a(2a - 3b)$

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Thus, the factorised form of $14a^2 - 21ab$ is $7a(2a - 3b)$.

Given: The algebraic expression is $3x^2y - 6xy^2 + 9xy$.

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To Factorise: The expression $3x^2y - 6xy^2 + 9xy$ by taking out the common factors.

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Solution:

The given expression has three terms: $3x^2y$, $-6xy^2$, and $9xy$.

We need to find the greatest common factor (GCF) of all these terms.

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First, let's find the GCF of the numerical coefficients: 3, -6, and 9.

The absolute values of the coefficients are 3, 6, and 9.

The factors of 3 are 1, 3.

The factors of 6 are 1, 2, 3, 6.

The factors of 9 are 1, 3, 9.

The greatest common factor of 3, 6, and 9 is 3.

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Next, let's find the GCF of the variable parts: $x^2y$, $xy^2$, and $xy$.

Term 1: $x^2y = x \times x \times y$. The power of $x$ is 2, and the power of $y$ is 1.

Term 2: $xy^2 = x \times y \times y$. The power of $x$ is 1, and the power of $y$ is 2.

Term 3: $xy = x \times y$. The power of $x$ is 1, and the power of $y$ is 1.

To find the GCF of the variables, we take the lowest power of each common variable present in all terms.

The variable $x$ is present in all terms. The lowest power of $x$ is $x^1$.

The variable $y$ is present in all terms. The lowest power of $y$ is $y^1$.

The GCF of the variable parts is $xy$.

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The GCF of the entire expression is the product of the GCF of the numerical coefficients and the GCF of the variable parts.

GCF $= 3 \times xy = 3xy$.

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Now, we rewrite each term of the expression by factoring out the GCF, $3xy$:

$3x^2y = 3xy \times x$

$-6xy^2 = 3xy \times (-2y)$

$9xy = 3xy \times 3$

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Substitute these back into the expression:

$3x^2y - 6xy^2 + 9xy = (3xy \times x) + (3xy \times (-2y)) + (3xy \times 3)$

Now, take out the common factor $3xy$:

$3x^2y - 6xy^2 + 9xy = 3xy(x - 2y + 3)$

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Thus, the factorised form of $3x^2y - 6xy^2 + 9xy$ is $3xy(x - 2y + 3)$.

Given: The algebraic expression is $ax + ay + bx + by$.

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To Factorise: The expression $ax + ay + bx + by$ by grouping terms.

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Solution:

The given expression has four terms. We can group the terms in pairs and look for common factors in each pair.

Let's group the first two terms and the last two terms:

$(ax + ay) + (bx + by)$

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Now, find the common factor in the first group $(ax + ay)$.

The common factor is $a$.

$ax + ay = a(x + y)$

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Next, find the common factor in the second group $(bx + by)$.

The common factor is $b$.

$bx + by = b(x + y)$

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Substitute these back into the grouped expression:

$a(x + y) + b(x + y)$

Now, we can see that $(x + y)$ is a common binomial factor in both terms.

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Factor out the common binomial factor $(x + y)$:

$a(x + y) + b(x + y) = (x + y)(a + b)$

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Thus, the factorised form of $ax + ay + bx + by$ is $(x + y)(a + b)$.

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Alternate Solution:

We can also group the terms differently.

Group the first and third terms, and the second and fourth terms:

$(ax + bx) + (ay + by)$

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Find the common factor in the first group $(ax + bx)$.

The common factor is $x$.

$ax + bx = x(a + b)$

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Find the common factor in the second group $(ay + by)$.

The common factor is $y$.

$ay + by = y(a + b)$

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Substitute these back into the grouped expression:

$x(a + b) + y(a + b)$

Now, we can see that $(a + b)$ is a common binomial factor in both terms.

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Factor out the common binomial factor $(a + b)$:

$x(a + b) + y(a + b) = (a + b)(x + y)$

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Both methods lead to the same factorised form: $(x + y)(a + b)$. The order of the factors does not matter.

Given: The algebraic expression is $pq - qr + ps - rs$.

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To Factorise: The expression $pq - qr + ps - rs$ by grouping terms.

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Solution:

The given expression has four terms. We will group the terms into two pairs and factor out the common factor from each pair.

Let's group the first two terms and the last two terms:

$(pq - qr) + (ps - rs)$

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In the first group, $(pq - qr)$, the common factor is $q$.

$pq - qr = q(p - r)$

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In the second group, $(ps - rs)$, the common factor is $s$.

$ps - rs = s(p - r)$

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Substitute these factorised groups back into the expression:

$q(p - r) + s(p - r)$

Now, we can see that $(p - r)$ is a common binomial factor in both terms.

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Factor out the common binomial factor $(p - r)$:

$q(p - r) + s(p - r) = (p - r)(q + s)$

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Thus, the factorised form of $pq - qr + ps - rs$ is $(p - r)(q + s)$.

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Alternate Solution:

We can also group the terms differently. Let's group the first and third terms, and the second and fourth terms:

$(pq + ps) + (-qr - rs)$

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In the first group, $(pq + ps)$, the common factor is $p$.

$pq + ps = p(q + s)$

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In the second group, $(-qr - rs)$, the common factor is $-r$.

$-qr - rs = -r(q + s)$

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Substitute these factorised groups back into the expression:

$p(q + s) + (-r)(q + s)$

$p(q + s) - r(q + s)$

Now, we can see that $(q + s)$ is a common binomial factor in both terms.

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Factor out the common binomial factor $(q + s)$:

$p(q + s) - r(q + s) = (q + s)(p - r)$

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Both methods yield the same result, $(p - r)(q + s)$ or $(q + s)(p - r)$, which is the factorised form of the given expression.

Given: The algebraic expression is $x^2 - 49$.

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To Factorise: The expression $x^2 - 49$ using the identity $a^2 - b^2$.

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Solution:

We need to factorise the expression $x^2 - 49$ by applying the difference of squares identity.

The identity for the difference of squares is:

$a^2 - b^2 = (a - b)(a + b)$

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We compare the given expression $x^2 - 49$ with the left side of the identity, $a^2 - b^2$.

The first term is $x^2$, which is already in the form of a square, where $a = x$.

The second term is $49$. We need to express $49$ as a square of some number.

We know that $7 \times 7 = 49$, so $49 = 7^2$.

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Now, we can rewrite the expression $x^2 - 49$ as:

$x^2 - 49 = x^2 - 7^2$

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This expression is now in the form $a^2 - b^2$, where $a = x$ and $b = 7$.

Applying the identity $a^2 - b^2 = (a - b)(a + b)$, we substitute $a=x$ and $b=7$:

$x^2 - 7^2 = (x - 7)(x + 7)$

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Thus, the factorised form of $x^2 - 49$ using the identity $a^2 - b^2$ is $(x - 7)(x + 7)$.

Given: The algebraic expression is $9a^2 - 16b^2$.

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To Factorise: The expression $9a^2 - 16b^2$ using an algebraic identity.

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Solution:

We observe that the given expression is a difference of two terms, where each term is a perfect square.

This suggests using the algebraic identity for the difference of squares:

$a^2 - b^2 = (a - b)(a + b)$

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Let's express each term in the given expression as a square.

The first term is $9a^2$. We can write $9$ as $3^2$, so $9a^2 = 3^2 \times a^2 = (3a)^2$.

The second term is $16b^2$. We can write $16$ as $4^2$, so $16b^2 = 4^2 \times b^2 = (4b)^2$.

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Now, substitute these square forms back into the expression:

$9a^2 - 16b^2 = (3a)^2 - (4b)^2$

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This expression is now in the form $A^2 - B^2$, where $A = 3a$ and $B = 4b$.

Applying the difference of squares identity, we replace $A$ with $3a$ and $B$ with $4b$:

$(3a)^2 - (4b)^2 = (3a - 4b)(3a + 4b)$

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Thus, the factorised form of $9a^2 - 16b^2$ is $(3a - 4b)(3a + 4b)$.

Given: The algebraic expression is $4x^2 - \frac{y^2}{9}$.

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To Factorise: The expression $4x^2 - \frac{y^2}{9}$ using an algebraic identity.

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Solution:

We need to factorise the given expression $4x^2 - \frac{y^2}{9}$ using an appropriate algebraic identity.

The expression is a difference of two terms. We observe that both terms are perfect squares.

The first term is $4x^2$. We can write $4x^2$ as $(2x)^2$, since $4 = 2^2$ and $x^2 = x^2$. So, $4x^2 = (2x)^2$.

The second term is $\frac{y^2}{9}$. We can write $\frac{y^2}{9}$ as $\left(\frac{y}{3}\right)^2$, since $y^2 = y^2$ and $9 = 3^2$. So, $\frac{y^2}{9} = \left(\frac{y}{3}\right)^2$.

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Now, we can rewrite the expression as:

$4x^2 - \frac{y^2}{9} = (2x)^2 - \left(\frac{y}{3}\right)^2$

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This expression is in the form of the difference of squares identity, which is:

$A^2 - B^2 = (A - B)(A + B)$

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Comparing $(2x)^2 - \left(\frac{y}{3}\right)^2$ with $A^2 - B^2$, we have $A = 2x$ and $B = \frac{y}{3}$.

Applying the identity, we substitute $A = 2x$ and $B = \frac{y}{3}$:

$(2x)^2 - \left(\frac{y}{3}\right)^2 = \left(2x - \frac{y}{3}\right)\left(2x + \frac{y}{3}\right)$

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Thus, the factorised form of $4x^2 - \frac{y^2}{9}$ is $\left(2x - \frac{y}{3}\right)\left(2x + \frac{y}{3}\right)$.

Given: The algebraic expression is $x^2 + 6x + 9$.

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To Factorise: The expression $x^2 + 6x + 9$ using the identity $(a+b)^2$.

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Solution:

We need to factorise the given expression $x^2 + 6x + 9$ by comparing it with the identity for the square of a binomial sum.

The identity is:

$(a+b)^2 = a^2 + 2ab + b^2$

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Let's examine the given expression $x^2 + 6x + 9$. It has three terms. We check if the first and last terms are perfect squares.

The first term is $x^2$. This can be written as $(x)^2$. So, we can take $a = x$.

The last term is $9$. This can be written as $3^2$. So, we can take $b = 3$.

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Now, let's check if the middle term, $6x$, matches $2ab$ using our values for $a$ and $b$.

$2ab = 2 \times (x) \times (3)$

$2ab = 6x$

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The calculated middle term ($6x$) matches the middle term in the given expression ($+6x$).

Since the expression $x^2 + 6x + 9$ is in the form $a^2 + 2ab + b^2$ with $a=x$ and $b=3$, we can factorise it as $(a+b)^2$.

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Substitute $a=x$ and $b=3$ into $(a+b)^2$:

$(x + 3)^2$

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Thus, the factorised form of $x^2 + 6x + 9$ using the identity $(a+b)^2$ is $(x + 3)^2$.

Given: The algebraic expression is $y^2 - 10y + 25$.

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To Factorise: The expression $y^2 - 10y + 25$ using the identity $(a-b)^2$.

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Solution:

We need to factorise the given expression $y^2 - 10y + 25$ by comparing it with the identity for the square of a binomial difference.

The identity is:

$(a-b)^2 = a^2 - 2ab + b^2$

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Let's examine the given expression $y^2 - 10y + 25$. It has three terms. We check if the first and last terms are perfect squares.

The first term is $y^2$. This can be written as $(y)^2$. So, we can take $a = y$.

The last term is $25$. This can be written as $5^2$. So, we can take $b = 5$.

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Now, let's check if the middle term, $-10y$, matches $-2ab$ using our values for $a$ and $b$.

$-2ab = -2 \times (y) \times (5)$

$-2ab = -10y$

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The calculated middle term ($-10y$) matches the middle term in the given expression ($-10y$).

Since the expression $y^2 - 10y + 25$ is in the form $a^2 - 2ab + b^2$ with $a=y$ and $b=5$, we can factorise it as $(a-b)^2$.

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Substitute $a=y$ and $b=5$ into $(a-b)^2$:

$(y - 5)^2$

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Thus, the factorised form of $y^2 - 10y + 25$ using the identity $(a-b)^2$ is $(y - 5)^2$.

Given: The trinomial is $x^2 + 7x + 10$.

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To Factorise: The trinomial $x^2 + 7x + 10$ by splitting the middle term.

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Solution:

The given trinomial is of the form $ax^2 + bx + c$, where $a=1$, $b=7$, and $c=10$.

To factorise by splitting the middle term, we need to find two numbers whose sum is $b$ (the coefficient of $x$) and whose product is $ac$ (the product of the coefficient of $x^2$ and the constant term).

Here, the sum is $b=7$ and the product is $ac = 1 \times 10 = 10$.

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We look for two numbers whose product is 10 and whose sum is 7.

Let's list pairs of factors of 10 and check their sums:

Factors of 10	Sum of factors
1 and 10	$1 + 10 = 11$
2 and 5	$2 + 5 = 7$
-1 and -10	$-1 + (-10) = -11$
-2 and -5	$-2 + (-5) = -7$

The pair of numbers that satisfy the conditions (product is 10, sum is 7) is 2 and 5.

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Now, we split the middle term $7x$ into $2x + 5x$:

$x^2 + 7x + 10 = x^2 + 2x + 5x + 10$

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Next, we group the terms in pairs:

$(x^2 + 2x) + (5x + 10)$

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Factor out the common factor from each group:

From the first group $(x^2 + 2x)$, the common factor is $x$.

$x^2 + 2x = x(x + 2)$

From the second group $(5x + 10)$, the common factor is 5.

$5x + 10 = 5(x + 2)$

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Substitute these factorised groups back into the expression:

$x(x + 2) + 5(x + 2)$

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Now, we observe that $(x + 2)$ is a common binomial factor in both terms. Factor out $(x + 2)$:

$x(x + 2) + 5(x + 2) = (x + 2)(x + 5)$

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Thus, the factorised form of the trinomial $x^2 + 7x + 10$ by splitting the middle term is $(x + 2)(x + 5)$.

Given: The trinomial is $y^2 - 5y + 6$.

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To Factorise: The trinomial $y^2 - 5y + 6$ by splitting the middle term.

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Solution:

The given trinomial is of the form $ay^2 + by + c$, where $a=1$, $b=-5$, and $c=6$.

To factorise by splitting the middle term, we need to find two numbers whose sum is $b$ (the coefficient of $y$) and whose product is $ac$ (the product of the coefficient of $y^2$ and the constant term).

Here, the sum is $b=-5$ and the product is $ac = 1 \times 6 = 6$.

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We look for two numbers whose product is 6 and whose sum is -5.

Since the product is positive and the sum is negative, both numbers must be negative.

Let's list pairs of negative factors of 6 and check their sums:

Negative Factors of 6	Sum of factors
-1 and -6	$-1 + (-6) = -7$
-2 and -3	$-2 + (-3) = -5$

The pair of numbers that satisfy the conditions (product is 6, sum is -5) is -2 and -3.

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Now, we split the middle term $-5y$ into $-2y - 3y$:

$y^2 - 5y + 6 = y^2 - 2y - 3y + 6$

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Next, we group the terms in pairs:

$(y^2 - 2y) + (-3y + 6)$

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Factor out the common factor from each group:

From the first group $(y^2 - 2y)$, the common factor is $y$.

$y^2 - 2y = y(y - 2)$

From the second group $(-3y + 6)$, the common factor is -3.

$-3y + 6 = -3(y - 2)$

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Substitute these factorised groups back into the expression:

$y(y - 2) - 3(y - 2)$

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Now, we observe that $(y - 2)$ is a common binomial factor in both terms. Factor out $(y - 2)$:

$y(y - 2) - 3(y - 2) = (y - 2)(y - 3)$

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Thus, the factorised form of the trinomial $y^2 - 5y + 6$ by splitting the middle term is $(y - 2)(y - 3)$.

Given: The polynomial is $18x^2y - 27xy^2$ and the monomial divisor is $9xy$.

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To Divide: $(18x^2y - 27xy^2) \div (9xy)$.

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Solution:

To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial.

The expression can be written as:

$\frac{18x^2y - 27xy^2}{9xy}$

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We can split this fraction into two separate fractions, one for each term in the numerator:

$\frac{18x^2y}{9xy} - \frac{27xy^2}{9xy}$

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Now, we divide each term separately.

For the first term $\frac{18x^2y}{9xy}$:

Divide the numerical coefficients: $\frac{18}{9} = 2$.

Divide the variable $x$: $\frac{x^2}{x} = x^{2-1} = x^1 = x$.

Divide the variable $y$: $\frac{y}{y} = y^{1-1} = y^0 = 1$.

So, $\frac{18x^2y}{9xy} = 2 \times x \times 1 = 2x$.

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For the second term $\frac{27xy^2}{9xy}$:

Divide the numerical coefficients: $\frac{27}{9} = 3$.

Divide the variable $x$: $\frac{x}{x} = x^{1-1} = x^0 = 1$.

Divide the variable $y$: $\frac{y^2}{y} = y^{2-1} = y^1 = y$.

So, $\frac{27xy^2}{9xy} = 3 \times 1 \times y = 3y$.

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Now, substitute the results back into the expression:

$\frac{18x^2y - 27xy^2}{9xy} = 2x - 3y$

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Alternatively, we can factor out the common factor from the polynomial in the numerator first.

The terms are $18x^2y$ and $27xy^2$.

GCF of 18 and 27 is 9.

GCF of $x^2y$ and $xy^2$ is $xy$.

So, the GCF of the polynomial is $9xy$.

$18x^2y - 27xy^2 = 9xy(2x - 3y)$

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Now, perform the division:

$\frac{9xy(2x - 3y)}{9xy}$

Cancel the common factor $9xy$ from the numerator and the denominator:

$\frac{\cancel{9xy}(2x - 3y)}{\cancel{9xy}} = 2x - 3y$

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Thus, the result of the division is $2x - 3y$.

Given: The polynomial is $p^3 - 4p^2 + 6p$ and the monomial divisor is $p$.

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To Divide: $(p^3 - 4p^2 + 6p) \div p$.

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Solution:

To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial divisor.

The expression can be written as:

$\frac{p^3 - 4p^2 + 6p}{p}$

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We can rewrite this expression by dividing each term in the numerator by the denominator:

$\frac{p^3}{p} - \frac{4p^2}{p} + \frac{6p}{p}$

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Now, we simplify each term using the rule of exponents for division, $\frac{a^m}{a^n} = a^{m-n}$ (for $a \neq 0$).

For the first term $\frac{p^3}{p}$:

$\frac{p^3}{p} = p^{3-1} = p^2$

For the second term $\frac{4p^2}{p}$:

$\frac{4p^2}{p} = 4 \times \frac{p^2}{p} = 4 \times p^{2-1} = 4p^1 = 4p$

For the third term $\frac{6p}{p}$:

$\frac{6p}{p} = 6 \times \frac{p}{p} = 6 \times p^{1-1} = 6 \times p^0 = 6 \times 1 = 6$

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Now, combine the simplified terms:

$p^2 - 4p + 6$

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Alternate Solution:

We can also factor out the common monomial factor from the polynomial in the numerator.

The terms in the polynomial $p^3 - 4p^2 + 6p$ are $p^3$, $-4p^2$, and $6p$.

The common factor among these terms is $p$.

$p^3 - 4p^2 + 6p = p(p^2 - 4p + 6)$

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Now, perform the division:

$\frac{p(p^2 - 4p + 6)}{p}$

Cancel the common factor $p$ from the numerator and the denominator (assuming $p \neq 0$):

$\frac{\cancel{p}(p^2 - 4p + 6)}{\cancel{p}} = p^2 - 4p + 6$

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Both methods yield the same result.

Thus, the result of the division is $p^2 - 4p + 6$.

Given: The expression is $\frac{28a^4b^3}{7a^2b}$.

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To Simplify: The given expression by division.

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Solution:

The given expression involves dividing a monomial by a monomial.

We can separate the numerical and variable parts for division:

$\frac{28a^4b^3}{7a^2b} = \left(\frac{28}{7}\right) \times \left(\frac{a^4}{a^2}\right) \times \left(\frac{b^3}{b}\right)$

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Now, simplify each part:

1. Simplify the numerical part: $\frac{28}{7}$

$\frac{28}{7} = 4$

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2. Simplify the variable part with $a$: $\frac{a^4}{a^2}$

Using the rule of exponents $\frac{x^m}{x^n} = x^{m-n}$:

$\frac{a^4}{a^2} = a^{4-2} = a^2$

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3. Simplify the variable part with $b$: $\frac{b^3}{b}$

Using the rule of exponents $\frac{x^m}{x^n} = x^{m-n}$, noting that $b = b^1$:

$\frac{b^3}{b^1} = b^{3-1} = b^2$

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Now, combine the simplified parts:

$\frac{28a^4b^3}{7a^2b} = 4 \times a^2 \times b^2$

$\frac{28a^4b^3}{7a^2b} = 4a^2b^2$

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Alternatively, we can cancel common factors directly:

$\frac{28a^4b^3}{7a^2b} = \frac{\cancel{28}^{4} \times a^{4} \times b^{3}}{\cancel{7}_{1} \times a^{2} \times b^{1}}$

$\frac{\cancel{28}^{4} \times a^{\cancel{4}^{2}} \times b^{\cancel{3}^{2}}}{\cancel{7}_{1} \times \cancel{a^{2}}_{1} \times \cancel{b^{1}}_{1}} = 4 \times a^2 \times b^2 = 4a^2b^2$

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Thus, the simplified result of the division is $4a^2b^2$.

Given: The expression is $12x^2 - 27y^2$.

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To Factorise: The expression $12x^2 - 27y^2$ completely.

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Solution:

We are asked to factorise the expression $12x^2 - 27y^2$.

First, look for any common factors in the terms $12x^2$ and $-27y^2$.

The numerical coefficients are 12 and 27. Let's find their greatest common factor (GCF).

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 27: 1, 3, 9, 27

The GCF of 12 and 27 is 3.

There are no common variable factors (one term has $x^2$ and the other has $y^2$).

So, the common factor for the entire expression is 3.

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Factor out the common factor 3 from both terms:

$12x^2 - 27y^2 = 3(4x^2 - 9y^2)$

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Now, we need to factorise the expression inside the parentheses, which is $4x^2 - 9y^2$.

This expression is a difference of two terms. We check if they are perfect squares.

The first term is $4x^2$. This can be written as $(2x)^2$, since $4=2^2$ and $x^2=x^2$.

The second term is $9y^2$. This can be written as $(3y)^2$, since $9=3^2$ and $y^2=y^2$.

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So, $4x^2 - 9y^2 = (2x)^2 - (3y)^2$.

This is in the form of the difference of squares identity, $A^2 - B^2 = (A - B)(A + B)$, where $A = 2x$ and $B = 3y$.

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Applying the identity to $(2x)^2 - (3y)^2$:

$(2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y)$

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Now, substitute this back into the expression with the common factor we took out initially:

$12x^2 - 27y^2 = 3(4x^2 - 9y^2) = 3(2x - 3y)(2x + 3y)$

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The expression is now completely factorised.

Thus, the factorised form of $12x^2 - 27y^2$ is $3(2x - 3y)(2x + 3y)$.

Given: The expression is $5(x+y)^2 - 20(x+y)$.

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To Factorise: The expression $5(x+y)^2 - 20(x+y)$ completely.

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Solution:

The given expression has two terms: $5(x+y)^2$ and $-20(x+y)$.

We look for common factors in these two terms.

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Consider the numerical coefficients: 5 and -20. The greatest common factor (GCF) of 5 and 20 is 5.

Consider the binomial factor: $(x+y)$ is present in both terms.

In the first term, the factor is $(x+y)^2$.

In the second term, the factor is $(x+y)^1$.

The lowest power of the common binomial factor $(x+y)$ is $(x+y)^1$.

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The greatest common factor (GCF) of the entire expression is the product of the common numerical factor and the common binomial factor with the lowest power.

GCF $= 5(x+y)$.

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Now, we factor out the GCF $5(x+y)$ from each term of the expression.

The first term is $5(x+y)^2$. Dividing this by the GCF $5(x+y)$, we get:

$\frac{5(x+y)^2}{5(x+y)} = \frac{\cancel{5}\cancel{(x+y)}^2}{\cancel{5}\cancel{(x+y)}} = (x+y)^{2-1} = (x+y)^1 = x+y$

The second term is $-20(x+y)$. Dividing this by the GCF $5(x+y)$, we get:

$\frac{-20(x+y)}{5(x+y)} = \frac{\cancel{-20}^{-4}\cancel{(x+y)}}{\cancel{5}_{1}\cancel{(x+y)}} = -4$

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So, we can write the original expression as the GCF multiplied by the sum/difference of the results from the division:

$5(x+y)^2 - 20(x+y) = 5(x+y) \left(\frac{5(x+y)^2}{5(x+y)} + \frac{-20(x+y)}{5(x+y)}\right)$

$5(x+y)^2 - 20(x+y) = 5(x+y) (x+y - 4)$

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The expression is now completely factorised as a product of a numerical factor (5), a binomial factor $(x+y)$, and another binomial factor $(x+y-4)$.

Thus, the factorised form of $5(x+y)^2 - 20(x+y)$ is $5(x+y)(x+y - 4)$.

Given: The polynomial is $x^2 + 5x + 6$ and the divisor is $x+2$.

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To Divide: $(x^2 + 5x + 6) \div (x+2)$.

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Solution (Polynomial Long Division):

We will perform polynomial long division to divide $x^2 + 5x + 6$ by $x+2$.

$\begin{array}{r} x+3\phantom{)} \\ x+2{\overline{\smash{\big)}\,x^2+5x+6\phantom{)}}} \\ \underline{-~\phantom{(}(x^2+2x)\phantom{+6)}} \\ 0+3x+6\phantom{)} \\ \underline{-~\phantom{()}(3x+6)} \\ 0\phantom{)} \end{array}$

Step 1: Divide the leading term of the dividend ($x^2$) by the leading term of the divisor ($x$). $x^2 \div x = x$. Write $x$ in the quotient.

Step 2: Multiply the divisor ($x+2$) by the term just placed in the quotient ($x$). $x(x+2) = x^2 + 2x$. Write this below the dividend.

Step 3: Subtract $(x^2 + 2x)$ from $(x^2 + 5x + 6)$. $(x^2 + 5x + 6) - (x^2 + 2x) = 3x + 6$. Bring down the next term (+6).

Step 4: Now, the new dividend is $3x + 6$. Divide the leading term of this new dividend ($3x$) by the leading term of the divisor ($x$). $3x \div x = 3$. Write +3 in the quotient.

Step 5: Multiply the divisor ($x+2$) by the new term in the quotient (3). $3(x+2) = 3x + 6$. Write this below $3x+6$.

Step 6: Subtract $(3x + 6)$ from $(3x + 6)$. $(3x + 6) - (3x + 6) = 0$. The remainder is 0.

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The quotient is $x+3$ and the remainder is 0.

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Alternate Solution (Factoring):

We can factor the dividend $x^2 + 5x + 6$ and cancel out the common factor if it exists.

The trinomial $x^2 + 5x + 6$ is of the form $ax^2 + bx + c$, where $a=1, b=5, c=6$.

We look for two numbers that multiply to $ac = 1 \times 6 = 6$ and add up to $b = 5$. The numbers are 2 and 3.

We split the middle term $5x$ into $2x + 3x$:

$x^2 + 5x + 6 = x^2 + 2x + 3x + 6$

Group the terms:

$(x^2 + 2x) + (3x + 6)$

Factor out common factors from each group:

$x(x + 2) + 3(x + 2)$

Factor out the common binomial factor $(x + 2)$:

$(x + 2)(x + 3)$

So, the dividend $x^2 + 5x + 6$ factorises to $(x + 2)(x + 3)$.

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Now, perform the division:

$\frac{x^2 + 5x + 6}{x+2} = \frac{(x+2)(x+3)}{x+2}$

Assuming the divisor $x+2 \neq 0$, we can cancel the common factor $(x+2)$ from the numerator and the denominator:

$\frac{\cancel{(x+2)}(x+3)}{\cancel{(x+2)}} = x+3$

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Both methods give the same result.

Thus, the result of the division $(x^2 + 5x + 6) \div (x+2)$ is $x+3$.

Given: The polynomial is $a^2 - 81$ and the divisor is $a - 9$.

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To Divide: $(a^2 - 81) \div (a - 9)$.

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Solution (Using Algebraic Identity):

We can factorise the dividend $a^2 - 81$ using the difference of squares identity.

The identity is $x^2 - y^2 = (x - y)(x + y)$.

The expression $a^2 - 81$ is in the form of a difference of squares, where $a^2$ is the square of $a$, and $81$ is the square of $9$ (since $9^2 = 81$).

So, we can write $a^2 - 81 = a^2 - 9^2$.

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Applying the difference of squares identity with $x=a$ and $y=9$:

$a^2 - 9^2 = (a - 9)(a + 9)$

So, the dividend $a^2 - 81$ can be factorised as $(a - 9)(a + 9)$.

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Now, we can perform the division:

$\frac{a^2 - 81}{a - 9} = \frac{(a - 9)(a + 9)}{a - 9}$

Assuming the divisor $a - 9 \neq 0$, we can cancel the common factor $(a - 9)$ from the numerator and the denominator:

$\frac{\cancel{(a - 9)}(a + 9)}{\cancel{(a - 9)}} = a + 9$

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Alternate Solution (Polynomial Long Division):

We can also perform polynomial long division to divide $a^2 - 81$ by $a - 9$. We write the dividend as $a^2 + 0a - 81$ to account for the missing $a$ term.

$\begin{array}{r} a+9\phantom{)} \\ a-9{\overline{\smash{\big)}\,a^2+0a-81\phantom{)}}} \\ \underline{-~\phantom{(}(a^2-9a)\phantom{-81)}} \\ 0+9a-81\phantom{)} \\ \underline{-~\phantom{()}(9a-81)} \\ 0\phantom{)} \end{array}$

Step 1: Divide $a^2$ by $a$ to get $a$. Write $a$ in the quotient.

Step 2: Multiply $a(a-9) = a^2 - 9a$. Subtract this from the dividend: $(a^2+0a-81) - (a^2 - 9a) = 9a - 81$.

Step 3: Divide $9a$ by $a$ to get $9$. Write $+9$ in the quotient.

Step 4: Multiply $9(a-9) = 9a - 81$. Subtract this from $9a - 81$: $(9a - 81) - (9a - 81) = 0$.

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The quotient is $a+9$ and the remainder is 0.

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Both methods give the same result.

Thus, the result of the division $(a^2 - 81) \div (a - 9)$ is $a + 9$.

Given: The algebraic expression is $4x^2 + 12xy + 9y^2$.

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To Factorise: The expression $4x^2 + 12xy + 9y^2$.

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Solution:

The given expression is a trinomial. We observe that the first term $4x^2$ and the last term $9y^2$ are perfect squares, and the middle term is positive. This suggests using the identity for the square of a binomial sum:

$(a+b)^2 = a^2 + 2ab + b^2$

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Let's try to express the given trinomial in the form $a^2 + 2ab + b^2$.

The first term is $4x^2$. We can write $4x^2$ as $(2x)^2$, since $4 = 2^2$ and $x^2 = x^2$. So, we can take $a = 2x$.

The last term is $9y^2$. We can write $9y^2$ as $(3y)^2$, since $9 = 3^2$ and $y^2 = y^2$. So, we can take $b = 3y$.

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Now, let's check if the middle term, $12xy$, matches $2ab$ using our values for $a$ and $b$.

$2ab = 2 \times (2x) \times (3y)$

$2ab = 4x \times 3y$

$2ab = 12xy$

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The calculated middle term ($12xy$) matches the middle term in the given expression ($+12xy$).

Since the expression $4x^2 + 12xy + 9y^2$ is in the form $a^2 + 2ab + b^2$ with $a=2x$ and $b=3y$, we can factorise it as $(a+b)^2$.

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Substitute $a=2x$ and $b=3y$ into $(a+b)^2$:

$(2x + 3y)^2$

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Thus, the factorised form of $4x^2 + 12xy + 9y^2$ is $(2x + 3y)^2$.

Given: The algebraic expression is $1 - 4m + 4m^2$.

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To Factorise: The expression $1 - 4m + 4m^2$.

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Solution:

The given expression is a trinomial. Let's rearrange the terms in descending powers of $m$ to make it easier to compare with standard identities:

$4m^2 - 4m + 1$

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We observe that the first term $4m^2$ and the last term $1$ are perfect squares, and the middle term has a negative sign. This suggests using the identity for the square of a binomial difference:

$(a-b)^2 = a^2 - 2ab + b^2$

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Let's try to express the trinomial $4m^2 - 4m + 1$ in the form $a^2 - 2ab + b^2$.

The first term is $4m^2$. We can write $4m^2$ as $(2m)^2$, since $4 = 2^2$ and $m^2 = m^2$. So, we can take $a = 2m$.

The last term is $1$. We can write $1$ as $1^2$. So, we can take $b = 1$.

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Now, let's check if the middle term, $-4m$, matches $-2ab$ using our values for $a$ and $b$.

$-2ab = -2 \times (2m) \times (1)$

$-2ab = -4m$

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The calculated middle term ($-4m$) matches the middle term in the rearranged expression ($-4m$).

Since the expression $4m^2 - 4m + 1$ is in the form $a^2 - 2ab + b^2$ with $a=2m$ and $b=1$, we can factorise it as $(a-b)^2$.

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Substitute $a=2m$ and $b=1$ into $(a-b)^2$:

$(2m - 1)^2$

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Thus, the factorised form of $1 - 4m + 4m^2$ (or $4m^2 - 4m + 1$) is $(2m - 1)^2$.

Given: The polynomial is $y^2 - 2y - 15$ and the divisor is $y+3$.

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To Divide: $(y^2 - 2y - 15) \div (y+3)$.

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Solution (Polynomial Long Division):

We will perform polynomial long division to divide $y^2 - 2y - 15$ by $y+3$.

$\begin{array}{r} y-5\phantom{)} \\ y+3{\overline{\smash{\big)}\,y^2-2y-15\phantom{)}}} \\ \underline{-~\phantom{(}(y^2+3y)\phantom{-15)}} \\ 0-5y-15\phantom{)} \\ \underline{-~\phantom{()}(-5y-15)} \\ 0\phantom{)} \end{array}$

Step 1: Divide the leading term of the dividend ($y^2$) by the leading term of the divisor ($y$). $y^2 \div y = y$. Write $y$ in the quotient.

Step 2: Multiply the divisor ($y+3$) by the term just placed in the quotient ($y$). $y(y+3) = y^2 + 3y$. Write this below the dividend.

Step 3: Subtract $(y^2 + 3y)$ from $(y^2 - 2y - 15)$. $(y^2 - 2y - 15) - (y^2 + 3y) = y^2 - 2y - 15 - y^2 - 3y = -5y - 15$. Bring down the next term (-15).

Step 4: Now, the new dividend is $-5y - 15$. Divide the leading term of this new dividend ($-5y$) by the leading term of the divisor ($y$). $-5y \div y = -5$. Write -5 in the quotient.

Step 5: Multiply the divisor ($y+3$) by the new term in the quotient (-5). $-5(y+3) = -5y - 15$. Write this below $-5y-15$.

Step 6: Subtract $(-5y - 15)$ from $(-5y - 15)$. $(-5y - 15) - (-5y - 15) = 0$. The remainder is 0.

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The quotient is $y-5$ and the remainder is 0.

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Alternate Solution (Factoring):

We can factor the dividend $y^2 - 2y - 15$ and cancel out the common factor if it exists.

The trinomial $y^2 - 2y - 15$ is of the form $ay^2 + by + c$, where $a=1, b=-2, c=-15$.

We look for two numbers that multiply to $ac = 1 \times (-15) = -15$ and add up to $b = -2$. The numbers are -5 and 3.

We split the middle term $-2y$ into $-5y + 3y$:

$y^2 - 2y - 15 = y^2 - 5y + 3y - 15$

Group the terms:

$(y^2 - 5y) + (3y - 15)$

Factor out common factors from each group:

$y(y - 5) + 3(y - 5)$

Factor out the common binomial factor $(y - 5)$:

$(y - 5)(y + 3)$

So, the dividend $y^2 - 2y - 15$ factorises to $(y - 5)(y + 3)$.

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Now, perform the division:

$\frac{y^2 - 2y - 15}{y+3} = \frac{(y-5)(y+3)}{y+3}$

Assuming the divisor $y+3 \neq 0$, we can cancel the common factor $(y+3)$ from the numerator and the denominator:

$\frac{(y-5)\cancel{(y+3)}}{\cancel{(y+3)}} = y-5$

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Both methods give the same result.

Thus, the result of the division $(y^2 - 2y - 15) \div (y+3)$ is $y-5$.

Given: The algebraic expression is $m^2 - mn - 2m + 2n$.

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To Factorise: The expression $m^2 - mn - 2m + 2n$ by grouping terms.

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Solution:

The given expression has four terms. We will group the terms into two pairs and factor out the common factor from each pair.

Let's group the first two terms and the last two terms:

$(m^2 - mn) + (-2m + 2n)$

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In the first group, $(m^2 - mn)$, the common factor is $m$.

$m^2 - mn = m(m - n)$

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In the second group, $(-2m + 2n)$, the common factor is $-2$.

$-2m + 2n = -2(m - n)$

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Substitute these factorised groups back into the expression:

$m(m - n) + (-2)(m - n)$

$m(m - n) - 2(m - n)$

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Now, we can see that $(m - n)$ is a common binomial factor in both terms.

Factor out the common binomial factor $(m - n)$:

$m(m - n) - 2(m - n) = (m - n)(m - 2)$

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Thus, the factorised form of $m^2 - mn - 2m + 2n$ is $(m - n)(m - 2)$.

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Alternate Solution:

We can also group the terms differently. Let's group the first and third terms, and the second and fourth terms:

$(m^2 - 2m) + (-mn + 2n)$

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In the first group, $(m^2 - 2m)$, the common factor is $m$.

$m^2 - 2m = m(m - 2)$

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In the second group, $(-mn + 2n)$, the common factor is $-n$.

$-mn + 2n = -n(m - 2)$

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Substitute these factorised groups back into the expression:

$m(m - 2) + (-n)(m - 2)$

$m(m - 2) - n(m - 2)$

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Now, we can see that $(m - 2)$ is a common binomial factor in both terms.

Factor out the common binomial factor $(m - 2)$:

$m(m - 2) - n(m - 2) = (m - 2)(m - n)$

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Both grouping methods result in the same factorised form, $(m - n)(m - 2)$, or $(m - 2)(m - n)$.

(a) Factorise $18a^3b^2 - 27a^2b^3 + 36a^3b^3$

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Given: The expression is $18a^3b^2 - 27a^2b^3 + 36a^3b^3$.

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To Factorise: The expression completely by taking out common factors.

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Solution:

The given expression has three terms: $18a^3b^2$, $-27a^2b^3$, and $36a^3b^3$.

We find the greatest common factor (GCF) of these three terms.

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First, find the GCF of the numerical coefficients: 18, -27, and 36.

The absolute values are 18, 27, and 36.

Prime factorisation of 18: $2 \times 3^2$

Prime factorisation of 27: $3^3$

Prime factorisation of 36: $2^2 \times 3^2$

The common prime factor is 3, and the lowest power is $3^2$.

So, the GCF of 18, 27, and 36 is $3^2 = 9$.

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Next, find the GCF of the variable parts: $a^3b^2$, $a^2b^3$, and $a^3b^3$.

For variable $a$, the powers are 3, 2, and 3. The lowest power is $a^2$.

For variable $b$, the powers are 2, 3, and 3. The lowest power is $b^2$.

The GCF of the variable parts is $a^2b^2$.

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The overall GCF of the expression is the product of the numerical GCF and the variable GCF:

GCF $= 9 \times a^2b^2 = 9a^2b^2$.

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Now, factor out the GCF $9a^2b^2$ from each term:

$\frac{18a^3b^2}{9a^2b^2} = \frac{18}{9} \times \frac{a^3}{a^2} \times \frac{b^2}{b^2} = 2 \times a^{3-2} \times b^{2-2} = 2a^1b^0 = 2a$

$\frac{-27a^2b^3}{9a^2b^2} = \frac{-27}{9} \times \frac{a^2}{a^2} \times \frac{b^3}{b^2} = -3 \times a^{2-2} \times b^{3-2} = -3a^0b^1 = -3b$

$\frac{36a^3b^3}{9a^2b^2} = \frac{36}{9} \times \frac{a^3}{a^2} \times \frac{b^3}{b^2} = 4 \times a^{3-2} \times b^{3-2} = 4a^1b^1 = 4ab$

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Write the expression as the GCF multiplied by the sum/difference of the results:

$18a^3b^2 - 27a^2b^3 + 36a^3b^3 = 9a^2b^2(2a - 3b + 4ab)$

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The expression is now completely factorised.

The factorised form is $9a^2b^2(2a - 3b + 4ab)$.

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(b) Factorise $14xy(p+q) - 21x^2y(p+q)^2$

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Given: The expression is $14xy(p+q) - 21x^2y(p+q)^2$.

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To Factorise: The expression completely by taking out common factors.

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Solution:

The given expression has two terms: $14xy(p+q)$ and $-21x^2y(p+q)^2$.

We find the greatest common factor (GCF) of these two terms.

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First, find the GCF of the numerical coefficients: 14 and -21.

The absolute values are 14 and 21.

Prime factorisation of 14: $2 \times 7$

Prime factorisation of 21: $3 \times 7$

The common prime factor is 7. So, the GCF of 14 and 21 is 7.

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Next, find the GCF of the variable and binomial parts: $xy(p+q)$ and $x^2y(p+q)^2$.

For variable $x$, the powers are $x^1$ and $x^2$. The lowest power is $x^1 = x$.

For variable $y$, the powers are $y^1$ and $y^1$. The lowest power is $y^1 = y$.

For the binomial factor $(p+q)$, the powers are $(p+q)^1$ and $(p+q)^2$. The lowest power is $(p+q)^1 = (p+q)$.

The GCF of the variable and binomial parts is $xy(p+q)$.

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The overall GCF of the expression is the product of the numerical GCF and the variable/binomial GCF:

GCF $= 7 \times xy(p+q) = 7xy(p+q)$.

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Now, factor out the GCF $7xy(p+q)$ from each term:

$\frac{14xy(p+q)}{7xy(p+q)} = \frac{\cancel{14}^{2} \cancel{xy}\cancel{(p+q)}}{\cancel{7}_{1} \cancel{xy}\cancel{(p+q)}} = 2$

$\frac{-21x^2y(p+q)^2}{7xy(p+q)} = \frac{\cancel{-21}^{-3} x^{\cancel{2}^{1}}\cancel{y} (p+q)^{\cancel{2}^{1}}}{\cancel{7}_{1} \cancel{x}_{1}\cancel{y}_{1}\cancel{(p+q)}_{1}} = -3x(p+q)$

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Write the expression as the GCF multiplied by the difference of the results:

$14xy(p+q) - 21x^2y(p+q)^2 = 7xy(p+q)(2 - 3x(p+q))$

We can further simplify the term inside the last parenthesis by distributing -3x:

$14xy(p+q) - 21x^2y(p+q)^2 = 7xy(p+q)(2 - 3xp - 3xq)$

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The expression is now completely factorised.

The factorised form is $7xy(p+q)(2 - 3xp - 3xq)$.

(a) Factorise $xy - az + ay - xz$

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Given: The expression is $xy - az + ay - xz$.

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To Factorise: The expression $xy - az + ay - xz$ by grouping terms.

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Solution:

The given expression has four terms. We can rearrange and group the terms to find common factors.

Let's rearrange the terms first to group those with common variables:

$xy + ay - az - xz$

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Now, group the terms in pairs:

$(xy + ay) + (-az - xz)$

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Factor out the common factor from the first group $(xy + ay)$. The common factor is $y$.

$xy + ay = y(x + a)$

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Factor out the common factor from the second group $(-az - xz)$. The common factor is $-z$.

$-az - xz = -z(a + x)$

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Substitute these factorised groups back into the expression:

$y(x + a) + (-z)(a + x)$

$y(x + a) - z(a + x)$

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Since $(x + a)$ is the same as $(a + x)$ (due to the commutative property of addition), we can see that $(x + a)$ is a common binomial factor in both terms.

Factor out the common binomial factor $(x + a)$:

$y(x + a) - z(x + a) = (x + a)(y - z)$

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Thus, the factorised form of $xy - az + ay - xz$ is $(x + a)(y - z)$.

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Alternate Grouping:

We could also group the original terms as $(xy - xz) + (ay - az)$.

Factor out common factor from $(xy - xz)$: $x(y - z)$.

Factor out common factor from $(ay - az)$: $a(y - z)$.

The expression becomes $x(y - z) + a(y - z)$.

Factor out the common binomial $(y - z)$: $(y - z)(x + a)$.

This yields the same result.

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(b) Factorise $a^2 + ab + 9a + 9b$

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Given: The expression is $a^2 + ab + 9a + 9b$.

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To Factorise: The expression $a^2 + ab + 9a + 9b$ by grouping terms.

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Solution:

The given expression has four terms. We group the terms into pairs and factor out the common factor from each pair.

Group the first two terms and the last two terms:

$(a^2 + ab) + (9a + 9b)$

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In the first group, $(a^2 + ab)$, the common factor is $a$.

$a^2 + ab = a(a + b)$

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In the second group, $(9a + 9b)$, the common factor is 9.

$9a + 9b = 9(a + b)$

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Substitute these factorised groups back into the expression:

$a(a + b) + 9(a + b)$

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Now, we can see that $(a + b)$ is a common binomial factor in both terms.

Factor out the common binomial factor $(a + b)$:

$a(a + b) + 9(a + b) = (a + b)(a + 9)$

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Thus, the factorised form of $a^2 + ab + 9a + 9b$ is $(a + b)(a + 9)$.

(a) Factorise $p^4 - 256$

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Given: The expression is $p^4 - 256$.

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To Factorise: The expression $p^4 - 256$ using algebraic identities.

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Solution:

The given expression $p^4 - 256$ is a difference of two terms. We observe that both terms are perfect squares.

The first term is $p^4$, which can be written as $(p^2)^2$.

The second term is 256, which is $16^2$.

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So, we can rewrite the expression as $(p^2)^2 - 16^2$.

This is in the form of the difference of squares identity:

$A^2 - B^2 = (A - B)(A + B)$

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Comparing $(p^2)^2 - 16^2$ with $A^2 - B^2$, we have $A = p^2$ and $B = 16$.

Applying the identity:

$(p^2)^2 - 16^2 = (p^2 - 16)(p^2 + 16)$

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Now, we examine the factors obtained: $(p^2 - 16)$ and $(p^2 + 16)$.

The factor $(p^2 - 16)$ is also a difference of squares, since $p^2$ is the square of $p$, and $16$ is the square of $4$ ($4^2 = 16$).

Applying the difference of squares identity again to $(p^2 - 16)$ with $A=p$ and $B=4$:

$p^2 - 16 = p^2 - 4^2 = (p - 4)(p + 4)$

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The factor $(p^2 + 16)$ is a sum of squares and cannot be factorised further using real numbers.

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Substitute the factorisation of $(p^2 - 16)$ back into the expression:

$p^4 - 256 = (p^2 - 16)(p^2 + 16) = (p - 4)(p + 4)(p^2 + 16)$

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The expression is now completely factorised.

The factorised form of $p^4 - 256$ is $(p - 4)(p + 4)(p^2 + 16)$.

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(b) Factorise $9x^2 - 30xy + 25y^2$

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Given: The expression is $9x^2 - 30xy + 25y^2$.

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To Factorise: The expression $9x^2 - 30xy + 25y^2$ using algebraic identities.

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Solution:

The given expression is a trinomial. We check if it matches the identity for the square of a binomial difference:

$(a-b)^2 = a^2 - 2ab + b^2$

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We examine the terms of the expression $9x^2 - 30xy + 25y^2$.

The first term is $9x^2$. We can write $9x^2$ as $(3x)^2$, since $9=3^2$ and $x^2=x^2$. So, we can take $a = 3x$.

The last term is $25y^2$. We can write $25y^2$ as $(5y)^2$, since $25=5^2$ and $y^2=y^2$. So, we can take $b = 5y$.

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Now, we check if the middle term, $-30xy$, matches $-2ab$ using our values for $a$ and $b$.

$-2ab = -2 \times (3x) \times (5y)$

$-2ab = -6x \times 5y$

$-2ab = -30xy$

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The calculated middle term ($-30xy$) matches the middle term in the given expression ($-30xy$).

Since the expression $9x^2 - 30xy + 25y^2$ is in the form $a^2 - 2ab + b^2$ with $a=3x$ and $b=5y$, we can factorise it as $(a-b)^2$.

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Substitute $a=3x$ and $b=5y$ into $(a-b)^2$:

$(3x - 5y)^2$

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Thus, the factorised form of $9x^2 - 30xy + 25y^2$ is $(3x - 5y)^2$.

(a) Factorise $x^2 + 11x + 24$

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Given: The trinomial is $x^2 + 11x + 24$.

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To Factorise: The trinomial $x^2 + 11x + 24$ by splitting the middle term.

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Solution:

The trinomial is of the form $ax^2 + bx + c$, where $a=1$, $b=11$, and $c=24$.

We need to find two numbers whose sum is $b=11$ and whose product is $ac = 1 \times 24 = 24$.

We look for two numbers that multiply to 24 and add up to 11.

The pairs of factors of 24 are (1, 24), (2, 12), (3, 8), (4, 6).

Let's check the sum for each pair:

$1 + 24 = 25$

$2 + 12 = 14$

$3 + 8 = 11$

$4 + 6 = 10$

The numbers are 3 and 8, as their sum is 11 and their product is 24.

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Now, we split the middle term $11x$ into $3x + 8x$:

$x^2 + 11x + 24 = x^2 + 3x + 8x + 24$

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Next, we group the terms into pairs:

$(x^2 + 3x) + (8x + 24)$

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Factor out the common factor from each group:

From the first group $(x^2 + 3x)$, the common factor is $x$: $x(x + 3)$.

From the second group $(8x + 24)$, the common factor is 8: $8(x + 3)$.

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Substitute these factorised groups back into the expression:

$x(x + 3) + 8(x + 3)$

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Now, we see that $(x + 3)$ is a common binomial factor. Factor out $(x + 3)$:

$(x + 3)(x + 8)$

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Thus, the factorised form of $x^2 + 11x + 24$ is $(x + 3)(x + 8)$.

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(b) Factorise $a^2 - 10a + 21$

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Given: The trinomial is $a^2 - 10a + 21$.

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To Factorise: The trinomial $a^2 - 10a + 21$ by splitting the middle term.

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Solution:

The trinomial is of the form $aa^2 + ba + c$, where $a=1$, $b=-10$, and $c=21$.

We need to find two numbers whose sum is $b=-10$ and whose product is $ac = 1 \times 21 = 21$.

We look for two numbers that multiply to 21 and add up to -10.

Since the product is positive (21) and the sum is negative (-10), both numbers must be negative.

The pairs of negative factors of 21 are (-1, -21), (-3, -7).

Let's check the sum for each pair:

$-1 + (-21) = -22$

$-3 + (-7) = -10$

The numbers are -3 and -7, as their sum is -10 and their product is 21.

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Now, we split the middle term $-10a$ into $-3a - 7a$:

$a^2 - 10a + 21 = a^2 - 3a - 7a + 21$

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Next, we group the terms into pairs:

$(a^2 - 3a) + (-7a + 21)$

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Factor out the common factor from each group:

From the first group $(a^2 - 3a)$, the common factor is $a$: $a(a - 3)$.

From the second group $(-7a + 21)$, the common factor is -7: $-7(a - 3)$.

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Substitute these factorised groups back into the expression:

$a(a - 3) - 7(a - 3)$

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Now, we see that $(a - 3)$ is a common binomial factor. Factor out $(a - 3)$:

$(a - 3)(a - 7)$

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Thus, the factorised form of $a^2 - 10a + 21$ is $(a - 3)(a - 7)$.

(a) Factorise $y^2 - 3y - 18$

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Given: The trinomial is $y^2 - 3y - 18$.

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To Factorise: The trinomial $y^2 - 3y - 18$ by splitting the middle term.

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Solution:

The trinomial is of the form $ay^2 + by + c$, where $a=1$, $b=-3$, and $c=-18$.

To factorise by splitting the middle term, we need to find two numbers whose sum is $b=-3$ and whose product is $ac = 1 \times (-18) = -18$.

We look for two numbers that multiply to -18 and add up to -3.

Since the product is negative, one number must be positive and the other negative. Since the sum is negative, the negative number must have a larger absolute value.

Let's list pairs of factors of -18 and check their sums:

Factors of -18	Sum of factors
1 and -18	$1 + (-18) = -17$
2 and -9	$2 + (-9) = -7$
3 and -6	$3 + (-6) = -3$
-1 and 18	$-1 + 18 = 17$
-2 and 9	$-2 + 9 = 7$
-3 and 6	$-3 + 6 = 3$

The pair of numbers that satisfy the conditions (product is -18, sum is -3) is 3 and -6.

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Now, we split the middle term $-3y$ into $3y - 6y$:

$y^2 - 3y - 18 = y^2 + 3y - 6y - 18$

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Next, we group the terms in pairs:

$(y^2 + 3y) + (-6y - 18)$

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Factor out the common factor from each group:

From the first group $(y^2 + 3y)$, the common factor is $y$.

$y^2 + 3y = y(y + 3)$

From the second group $(-6y - 18)$, the common factor is -6.

$-6y - 18 = -6(y + 3)$

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Substitute these factorised groups back into the expression:

$y(y + 3) - 6(y + 3)$

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Now, we observe that $(y + 3)$ is a common binomial factor in both terms. Factor out $(y + 3)$:

$y(y + 3) - 6(y + 3) = (y + 3)(y - 6)$

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Thus, the factorised form of the trinomial $y^2 - 3y - 18$ by splitting the middle term is $(y + 3)(y - 6)$.

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(b) Factorise $m^2 + m - 30$

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Given: The trinomial is $m^2 + m - 30.

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To Factorise: The trinomial $m^2 + m - 30$ by splitting the middle term.

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Solution:

The trinomial is of the form $am^2 + bm + c$, where $a=1$, $b=1$, and $c=-30$.

To factorise by splitting the middle term, we need to find two numbers whose sum is $b=1$ and whose product is $ac = 1 \times (-30) = -30$.

We look for two numbers that multiply to -30 and add up to 1.

Since the product is negative, one number must be positive and the other negative. Since the sum is positive, the positive number must have a larger absolute value.

Let's list pairs of factors of -30 and check their sums:

Factors of -30	Sum of factors
1 and -30	$1 + (-30) = -29$
2 and -15	$2 + (-15) = -13$
3 and -10	$3 + (-10) = -7$
5 and -6	$5 + (-6) = -1$
6 and -5	$6 + (-5) = 1$

The pair of numbers that satisfy the conditions (product is -30, sum is 1) is 6 and -5.

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Now, we split the middle term $m$ into $6m - 5m$:

$m^2 + m - 30 = m^2 + 6m - 5m - 30$

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Next, we group the terms in pairs:

$(m^2 + 6m) + (-5m - 30)$

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Factor out the common factor from each group:

From the first group $(m^2 + 6m)$, the common factor is $m$.

$m^2 + 6m = m(m + 6)$

From the second group $(-5m - 30)$, the common factor is -5.

$-5m - 30 = -5(m + 6)$

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Substitute these factorised groups back into the expression:

$m(m + 6) - 5(m + 6)$

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Now, we observe that $(m + 6)$ is a common binomial factor in both terms. Factor out $(m + 6)$:

$m(m + 6) - 5(m + 6) = (m + 6)(m - 5)$

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Thus, the factorised form of the trinomial $m^2 + m - 30$ by splitting the middle term is $(m + 6)(m - 5)$.

Given: The polynomial (dividend) is $x^2 - 13x + 36$ and the binomial (divisor) is $x - 4$.

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To Divide: $(x^2 - 13x + 36) \div (x - 4)$ using polynomial long division.

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Solution (Polynomial Long Division):

We set up the long division as follows:

$\begin{array}{r} x-9\phantom{)} \\ x-4{\overline{\smash{\big)}\,x^2-13x+36\phantom{)}}} \\ \underline{-~\phantom{(}(x^2-4x)\phantom{+36)}} \\ 0-9x+36\phantom{)} \\ \underline{-~\phantom{()}(-9x+36)} \\ 0\phantom{)} \end{array}$

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Let's explain the steps of the long division:

Step 1: Divide the leading term of the dividend ($x^2$) by the leading term of the divisor ($x$).

$x^2 \div x = x$. Write this $x$ as the first term of the quotient.

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Step 2: Multiply the divisor ($x-4$) by the term just found in the quotient ($x$).

$x(x-4) = x^2 - 4x$. Write this result below the dividend, aligning like terms.

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Step 3: Subtract the result from the dividend. Remember to change the signs of the terms being subtracted.

$(x^2 - 13x + 36) - (x^2 - 4x) = x^2 - 13x + 36 - x^2 + 4x = (x^2 - x^2) + (-13x + 4x) + 36 = 0 - 9x + 36 = -9x + 36$.

Bring down the next term of the dividend (+36) if necessary. Here, it is already included in the subtraction.

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Step 4: Now, consider the new polynomial obtained from the subtraction ($-9x + 36$) as the new dividend. Divide its leading term ($-9x$) by the leading term of the divisor ($x$).

$-9x \div x = -9$. Write this $-9$ as the next term of the quotient.

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Step 5: Multiply the divisor ($x-4$) by the new term in the quotient ($-9$).

$-9(x-4) = -9x + 36$. Write this result below the new dividend ($-9x + 36$), aligning like terms.

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Step 6: Subtract the result from the new dividend.

$(-9x + 36) - (-9x + 36) = -9x + 36 + 9x - 36 = (-9x + 9x) + (36 - 36) = 0 + 0 = 0$.

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The remainder is 0.

The division is complete because the remainder is 0 (or its degree is less than the degree of the divisor).

The quotient is the expression on top of the division bar.

Quotient $= x - 9$

Remainder $= 0$

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Thus, the result of the division $(x^2 - 13x + 36) \div (x - 4)$ is $x - 9$.

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Alternate Solution (Factoring):

We can factor the dividend $x^2 - 13x + 36$ by splitting the middle term.

We need two numbers that multiply to $1 \times 36 = 36$ and add up to -13.

Since the product is positive and the sum is negative, both numbers must be negative. The numbers are -4 and -9 (since $(-4) \times (-9) = 36$ and $-4 + (-9) = -13$).

Split the middle term: $x^2 - 4x - 9x + 36$

Group terms: $(x^2 - 4x) + (-9x + 36)$

Factor each group: $x(x - 4) - 9(x - 4)$

Factor out the common binomial: $(x - 4)(x - 9)$

So, $x^2 - 13x + 36 = (x - 4)(x - 9)$.

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Now, perform the division:

$\frac{x^2 - 13x + 36}{x - 4} = \frac{(x - 4)(x - 9)}{x - 4}$

Assuming $x - 4 \neq 0$, we can cancel the common factor $(x - 4)$:

$\frac{\cancel{(x - 4)}(x - 9)}{\cancel{(x - 4)}} = x - 9$

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Both methods yield the same quotient.

Given: The polynomial (dividend) is $a^3 + 6a^2 + 12a + 8$ and the binomial (divisor) is $a + 2$.

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To Divide: $(a^3 + 6a^2 + 12a + 8) \div (a + 2)$ using polynomial long division.

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Solution (Polynomial Long Division):

We perform the polynomial long division as follows:

$\begin{array}{r} a^2+4a+4\phantom{)} \\ a+2{\overline{\smash{\big)}\,a^3+6a^2+12a+8\phantom{)}}} \\ \underline{-~\phantom{(}(a^3+2a^2)\phantom{+12a+8)}} \\ 0+4a^2+12a+8\phantom{)} \\ \underline{-~\phantom{()}(4a^2+8a)\phantom{+8)}} \\ 0+4a+8\phantom{)} \\ \underline{-~\phantom{()}(4a+8)} \\ 0\phantom{)} \end{array}$

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Let's detail the steps of the long division:

Step 1: Divide the leading term of the dividend ($a^3$) by the leading term of the divisor ($a$). $\frac{a^3}{a} = a^2$. Write $a^2$ as the first term of the quotient.

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Step 2: Multiply the divisor ($a+2$) by the term just found in the quotient ($a^2$). $a^2(a+2) = a^3 + 2a^2$. Write this result below the dividend, aligning like terms.

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Step 3: Subtract the result from the dividend: $(a^3 + 6a^2 + 12a + 8) - (a^3 + 2a^2) = 4a^2 + 12a + 8$. Bring down the remaining terms.

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Step 4: Consider $4a^2 + 12a + 8$ as the new dividend. Divide its leading term ($4a^2$) by the leading term of the divisor ($a$). $\frac{4a^2}{a} = 4a$. Write $+4a$ as the next term of the quotient.

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Step 5: Multiply the divisor ($a+2$) by the new term in the quotient ($4a$). $4a(a+2) = 4a^2 + 8a$. Write this result below $4a^2 + 12a + 8$, aligning like terms.

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Step 6: Subtract the result from $4a^2 + 12a + 8$: $(4a^2 + 12a + 8) - (4a^2 + 8a) = 4a + 8$. Bring down the remaining terms.

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Step 7: Consider $4a + 8$ as the new dividend. Divide its leading term ($4a$) by the leading term of the divisor ($a$). $\frac{4a}{a} = 4$. Write $+4$ as the next term of the quotient.

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Step 8: Multiply the divisor ($a+2$) by the new term in the quotient (4). $4(a+2) = 4a + 8$. Write this result below $4a + 8$, aligning like terms.

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Step 9: Subtract the result from $4a + 8$: $(4a + 8) - (4a + 8) = 0$.

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The remainder is 0.

The division is complete as the remainder is 0.

The quotient is the expression obtained on top of the division bar, which is $a^2 + 4a + 4$.

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Thus, the result of the division $(a^3 + 6a^2 + 12a + 8) \div (a + 2)$ is $a^2 + 4a + 4$.

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Alternate Solution (Factoring using Identity):

We can recognize that the dividend $a^3 + 6a^2 + 12a + 8$ is the expansion of the cube of a binomial $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Comparing $a^3 + 6a^2 + 12a + 8$ with $a^3 + 3a^2b + 3ab^2 + b^3$:

The first term $a^3$ matches.

The last term is $8$, which is $2^3$. So, we can assume $b=2$.

Let's check the middle terms with $b=2$:

$3a^2b = 3a^2(2) = 6a^2$ (Matches the second term)

$3ab^2 = 3a(2^2) = 3a(4) = 12a$ (Matches the third term)

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So, the dividend $a^3 + 6a^2 + 12a + 8$ is equal to $(a+2)^3$.

We can write $(a+2)^3$ as $(a+2)(a+2)^2$.

Now, perform the division:

$\frac{a^3 + 6a^2 + 12a + 8}{a + 2} = \frac{(a+2)^3}{a + 2} = \frac{(a+2)(a+2)^2}{a + 2}$

Assuming $a+2 \neq 0$, we can cancel the common factor $(a+2)$:

$\frac{\cancel{(a+2)}(a+2)^2}{\cancel{(a+2)}} = (a+2)^2$

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Expanding $(a+2)^2$ using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$(a+2)^2 = a^2 + 2(a)(2) + 2^2 = a^2 + 4a + 4$

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Both methods yield the same quotient: $a^2 + 4a + 4$.

Given: The expression is $\frac{(p^2 - 9q^2)}{(p - 3q)}$.

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To Simplify: The given expression by division, using an algebraic identity in the numerator.

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Solution:

The given expression is $\frac{p^2 - 9q^2}{p - 3q}$.

We need to factorise the numerator $p^2 - 9q^2$ using an algebraic identity.

The numerator $p^2 - 9q^2$ is a difference of two perfect squares.

We can write $p^2$ as $(p)^2$ and $9q^2$ as $(3q)^2$, since $9 = 3^2$ and $q^2 = q^2$.

So, the numerator can be written as $p^2 - (3q)^2$.

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We use the difference of squares identity:

$a^2 - b^2 = (a - b)(a + b)$

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Comparing $p^2 - (3q)^2$ with $a^2 - b^2$, we have $a = p$ and $b = 3q$.

Applying the identity to the numerator:

$p^2 - (3q)^2 = (p - 3q)(p + 3q)$

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Now, substitute this factorised form of the numerator back into the original expression:

$\frac{p^2 - 9q^2}{p - 3q} = \frac{(p - 3q)(p + 3q)}{(p - 3q)}$

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Assuming the denominator $p - 3q \neq 0$, we can cancel the common factor $(p - 3q)$ from the numerator and the denominator:

$\frac{\cancel{(p - 3q)}(p + 3q)}{\cancel{(p - 3q)}} = p + 3q$

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Thus, the simplified result of the division is $p + 3q$.

Given: The expression is $\frac{(4x^2 - 20x + 25)}{(2x - 5)}$.

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To Simplify: The given expression by division, using an algebraic identity in the numerator.

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Solution:

The given expression is $\frac{4x^2 - 20x + 25}{2x - 5}$.

We need to factorise the numerator $4x^2 - 20x + 25$ using an algebraic identity.

The numerator is a trinomial. We observe that the first term ($4x^2$) and the last term (25) are perfect squares, and the middle term is negative. This suggests the identity for the square of a binomial difference:

$(a-b)^2 = a^2 - 2ab + b^2$

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Let's express the terms of the numerator in the form of squares:

The first term is $4x^2$. We can write $4x^2$ as $(2x)^2$, since $4=2^2$ and $x^2=x^2$. So, we can take $a = 2x$.

The last term is $25$. We can write $25$ as $5^2$. So, we can take $b = 5$.

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Now, we check if the middle term, $-20x$, matches $-2ab$ using $a=2x$ and $b=5$.

$-2ab = -2 \times (2x) \times (5)$

$-2ab = -4x \times 5$

$-2ab = -20x$

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The calculated middle term ($-20x$) matches the middle term in the numerator ($-20x$).

Since the numerator $4x^2 - 20x + 25$ is in the form $a^2 - 2ab + b^2$ with $a=2x$ and $b=5$, we can factorise it as $(a-b)^2$.

So, $4x^2 - 20x + 25 = (2x - 5)^2$.

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Now, substitute this factorised form of the numerator back into the original expression:

$\frac{(2x - 5)^2}{(2x - 5)}$

We can write $(2x - 5)^2$ as $(2x - 5)(2x - 5)$.

$\frac{(2x - 5)(2x - 5)}{(2x - 5)}$

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Assuming the denominator $2x - 5 \neq 0$, we can cancel the common factor $(2x - 5)$ from the numerator and the denominator:

$\frac{\cancel{(2x - 5)}(2x - 5)}{\cancel{(2x - 5)}} = 2x - 5$

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Thus, the simplified result of the division is $2x - 5$.

Given: The expression is $x^4 - 16y^4$.

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To Factorise: The expression $x^4 - 16y^4$ completely.

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Solution:

We need to factorise the expression $x^4 - 16y^4$.

We observe that the expression is a difference of two terms, and both terms are perfect squares.

The first term is $x^4$, which can be written as $(x^2)^2$.

The second term is $16y^4$. We can write $16$ as $4^2$ and $y^4$ as $(y^2)^2$. So, $16y^4 = 4^2 \times (y^2)^2 = (4y^2)^2$.

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Now, we can rewrite the expression as the difference of squares:

$x^4 - 16y^4 = (x^2)^2 - (4y^2)^2$

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We use the algebraic identity for the difference of squares:

$a^2 - b^2 = (a - b)(a + b)$

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Comparing $(x^2)^2 - (4y^2)^2$ with $a^2 - b^2$, we have $a = x^2$ and $b = 4y^2$.

Applying the identity:

$(x^2)^2 - (4y^2)^2 = (x^2 - 4y^2)(x^2 + 4y^2)$

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Now, we examine the factors obtained: $(x^2 - 4y^2)$ and $(x^2 + 4y^2)$.

The factor $(x^2 + 4y^2)$ is a sum of squares and cannot be factorised further using real numbers.

The factor $(x^2 - 4y^2)$ is a difference of two terms, which are also perfect squares.

$x^2$ is the square of $x$, i.e., $(x)^2$.

$4y^2$ is the square of $2y$, i.e., $(2y)^2$, since $4=2^2$ and $y^2=y^2$.

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So, we can apply the difference of squares identity again to $(x^2 - 4y^2)$:

$x^2 - 4y^2 = (x)^2 - (2y)^2$

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=x$ and $b=2y$:

$x^2 - (2y)^2 = (x - 2y)(x + 2y)$

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Substitute this factorisation of $(x^2 - 4y^2)$ back into the previous result:

$x^4 - 16y^4 = (x^2 - 4y^2)(x^2 + 4y^2) = (x - 2y)(x + 2y)(x^2 + 4y^2)$

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The factors $(x - 2y)$, $(x + 2y)$, and $(x^2 + 4y^2)$ cannot be factorised further using real coefficients. Therefore, the factorisation is complete.

Thus, the factorised form of $x^4 - 16y^4$ is $(x - 2y)(x + 2y)(x^2 + 4y^2)$.

Given: The polynomial (dividend) is $x^3 - 6x^2 + 11x - 6$ and the binomial (divisor) is $x - 1$.

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To Divide: $(x^3 - 6x^2 + 11x - 6) \div (x - 1)$ using polynomial long division.

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Solution (Polynomial Long Division):

We perform the polynomial long division as follows:

$\begin{array}{r} x^2-5x+6\phantom{)} \\ x-1{\overline{\smash{\big)}\,x^3-6x^2+11x-6\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-x^2)\phantom{+11x-6)}} \\ 0-5x^2+11x-6\phantom{)} \\ \underline{-~\phantom{()}(-5x^2+5x)\phantom{-6)}} \\ 0+6x-6\phantom{)} \\ \underline{-~\phantom{()}(6x-6)} \\ 0\phantom{)} \end{array}$

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Let's detail the steps of the long division:

Step 1: Divide the leading term of the dividend ($x^3$) by the leading term of the divisor ($x$). $\frac{x^3}{x} = x^2$. Write $x^2$ as the first term of the quotient.

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Step 2: Multiply the divisor ($x-1$) by the term just found in the quotient ($x^2$). $x^2(x-1) = x^3 - x^2$. Write this result below the dividend.

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Step 3: Subtract the result from the dividend: $(x^3 - 6x^2 + 11x - 6) - (x^3 - x^2) = x^3 - 6x^2 + 11x - 6 - x^3 + x^2 = -5x^2 + 11x - 6$. Bring down the remaining terms.

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Step 4: Consider $-5x^2 + 11x - 6$ as the new dividend. Divide its leading term ($-5x^2$) by the leading term of the divisor ($x$). $\frac{-5x^2}{x} = -5x$. Write $-5x$ as the next term of the quotient.

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Step 5: Multiply the divisor ($x-1$) by the new term in the quotient ($-5x$). $-5x(x-1) = -5x^2 + 5x$. Write this result below $-5x^2 + 11x - 6$, aligning like terms.

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Step 6: Subtract the result from $-5x^2 + 11x - 6$: $(-5x^2 + 11x - 6) - (-5x^2 + 5x) = -5x^2 + 11x - 6 + 5x^2 - 5x = 6x - 6$. Bring down the remaining terms.

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Step 7: Consider $6x - 6$ as the new dividend. Divide its leading term ($6x$) by the leading term of the divisor ($x$). $\frac{6x}{x} = 6$. Write $+6$ as the next term of the quotient.

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Step 8: Multiply the divisor ($x-1$) by the new term in the quotient (6). $6(x-1) = 6x - 6$. Write this result below $6x - 6$, aligning like terms.

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Step 9: Subtract the result from $6x - 6$: $(6x - 6) - (6x - 6) = 0$.

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The remainder is 0.

The division is complete as the remainder is 0.

The quotient is the expression obtained on top of the division bar, which is $x^2 - 5x + 6$.

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Thus, the result of the division $(x^3 - 6x^2 + 11x - 6) \div (x - 1)$ is $x^2 - 5x + 6$.

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Alternate Solution (Factoring and Division):

We can attempt to factor the cubic polynomial $x^3 - 6x^2 + 11x - 6$. Since $(x-1)$ is a divisor and the remainder is 0 from long division, $(x-1)$ must be a factor.

The quotient obtained from the division is $x^2 - 5x + 6$. So, we can write:

$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$

Now, the division becomes:

$\frac{(x - 1)(x^2 - 5x + 6)}{(x - 1)}$

Assuming $x - 1 \neq 0$, we cancel the common factor:

$\frac{\cancel{(x - 1)}(x^2 - 5x + 6)}{\cancel{(x - 1)}} = x^2 - 5x + 6$

We can further factorise the quadratic $x^2 - 5x + 6$ by splitting the middle term. We need two numbers that multiply to $1 \times 6 = 6$ and add up to -5. The numbers are -2 and -3.

$x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = x(x-2) - 3(x-2) = (x-2)(x-3)$

So, the dividend $x^3 - 6x^2 + 11x - 6$ can be factorised as $(x-1)(x-2)(x-3)$.

The division can be written as:

$\frac{(x - 1)(x - 2)(x - 3)}{(x - 1)}$

Cancelling $(x-1)$, we get $(x-2)(x-3)$.

Expanding $(x-2)(x-3)$: $(x)(x) + (x)(-3) + (-2)(x) + (-2)(-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$.

Both methods confirm the quotient is $x^2 - 5x + 6$.

Given: The algebraic expression is $2a^2x + 4abx + 2b^2x - 81x$.

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To Factorise: The expression completely.

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Solution:

The given expression is $2a^2x + 4abx + 2b^2x - 81x$.

Following the hint, we first look for a common factor in all terms.

We can see that the variable $x$ is present in all four terms.

The numerical coefficients are 2, 4, 2, and -81. The greatest common factor of these numbers is 1. However, let's look at the first three terms: $2a^2x, 4abx, 2b^2x$. The common factor is $2x$. The last term is $-81x$. The common factor for the entire expression is $x$.

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Factor out the common factor $x$ from the entire expression:

$2a^2x + 4abx + 2b^2x - 81x = x(2a^2 + 4ab + 2b^2 - 81)$

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Now, we need to factorise the expression inside the parenthesis: $2a^2 + 4ab + 2b^2 - 81$.

Let's group the first three terms:

$(2a^2 + 4ab + 2b^2) - 81$

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Factor out the common numerical factor 2 from the first grouped terms:

$2(a^2 + 2ab + b^2) - 81$

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We recognise the expression inside the parenthesis $(a^2 + 2ab + b^2)$ as the expansion of the square of a binomial, $(a+b)^2$.

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, we substitute this into the expression:

$2(a+b)^2 - 81$

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Now, the expression is $2(a+b)^2 - 81$. We also notice that $81$ is a perfect square, $81 = 9^2$.

The expression is $2(a+b)^2 - 9^2$.

This expression is in the form of a difference of squares, $A^2 - B^2$, where $A^2 = 2(a+b)^2$ and $B^2 = 9^2$.

To fit the $A^2$ form, we can write $2(a+b)^2 = (\sqrt{2}(a+b))^2$.

So, the expression is $(\sqrt{2}(a+b))^2 - 9^2$.

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Now, we apply the difference of squares identity $A^2 - B^2 = (A - B)(A + B)$, where $A = \sqrt{2}(a+b)$ and $B = 9$.

$(\sqrt{2}(a+b))^2 - 9^2 = (\sqrt{2}(a+b) - 9)(\sqrt{2}(a+b) + 9)$

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Substitute this factorisation back into the expression we obtained after taking out the common factor $x$:

$x(2a^2 + 4ab + 2b^2 - 81) = x(\sqrt{2}(a+b) - 9)(\sqrt{2}(a+b) + 9)$

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Thus, the factorised form of $2a^2x + 4abx + 2b^2x - 81x$ is $x(\sqrt{2}(a+b) - 9)(\sqrt{2}(a+b) + 9)$.

Given: The trinomial is $6x^2 + 17x + 12$.

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To Factorise: The trinomial $6x^2 + 17x + 12$ by splitting the middle term.

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Solution:

The given trinomial is in the form $ax^2 + bx + c$, where $a=6$, $b=17$, and $c=12$.

To factorise by splitting the middle term, we need to find two numbers whose sum is $b$ (the coefficient of $x$) and whose product is $ac$ (the product of the coefficient of $x^2$ and the constant term).

Here, the sum is $b=17$ and the product is $ac = 6 \times 12 = 72$.

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We look for two numbers whose product is 72 and whose sum is 17.

Let's list pairs of factors of 72 and check their sums:

Factors of 72	Sum of factors
1 and 72	$1 + 72 = 73$
2 and 36	$2 + 36 = 38$
3 and 24	$3 + 24 = 27$
4 and 18	$4 + 18 = 22$
6 and 12	$6 + 12 = 18$
8 and 9	$8 + 9 = 17$

The pair of numbers that satisfy the conditions (product is 72, sum is 17) is 8 and 9.

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Now, we split the middle term $17x$ into $8x + 9x$ (or $9x + 8x$).

$6x^2 + 17x + 12 = 6x^2 + 8x + 9x + 12$

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Next, we group the terms in pairs:

$(6x^2 + 8x) + (9x + 12)$

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Factor out the common factor from each group:

From the first group $(6x^2 + 8x)$, the common factor is $2x$.

$6x^2 + 8x = 2x(3x + 4)$

From the second group $(9x + 12)$, the common factor is 3.

$9x + 12 = 3(3x + 4)$

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Substitute these factorised groups back into the expression:

$2x(3x + 4) + 3(3x + 4)$

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Now, we observe that $(3x + 4)$ is a common binomial factor in both terms. Factor out $(3x + 4)$:

$2x(3x + 4) + 3(3x + 4) = (3x + 4)(2x + 3)$

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Thus, the factorised form of the trinomial $6x^2 + 17x + 12$ by splitting the middle term is $(3x + 4)(2x + 3)$.