Chapter 12 Factorisation

Solution:

We need to factorise the expression $12a^2b + 15ab^2$ by finding the common factors among the terms.

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The given expression has two terms: $12a^2b$ and $15ab^2$.

Find the prime factors of the numerical coefficients:

$12 = 2 \times 2 \times 3$

$15 = 3 \times 5$

The common numerical factor is 3.

Look for common factors among the variables in each term:

The variable part of the first term is $a^2b = a \times a \times b$.

The variable part of the second term is $ab^2 = a \times b \times b$.

The common variables are $a$ and $b$. The lowest power of $a$ in both terms is $a^1 = a$. The lowest power of $b$ in both terms is $b^1 = b$.

The common variable factor is $ab$.

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The highest common factor (HCF) of the two terms is the product of the common numerical factor and the common variable factor.

HCF = $3 \times ab = 3ab$.

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Now, express each term as the product of the HCF and the remaining factor.

For the first term, $12a^2b = 3ab \times (4a)$.

For the second term, $15ab^2 = 3ab \times (5b)$.

Substitute these back into the original expression:

$12a^2b + 15ab^2 = 3ab \times (4a) + 3ab \times (5b)$

Using the distributive property, factor out the common factor $3ab$:

$= 3ab(4a + 5b)$

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The factorised form of $12a^2b + 15ab^2$ is $\mathbf{3ab(4a + 5b)}$.

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Check:

Multiply the factors to ensure they return the original expression:

$3ab(4a + 5b) = (3ab \times 4a) + (3ab \times 5b)$

$= 12a^2b + 15ab^2$

This matches the original expression.

The given expression is $10x^2 - 18x^3 + 14x^4$.

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We need to find the common factors among the terms $10x^2$, $-18x^3$, and $14x^4$.

The numerical coefficients are 10, -18, and 14. The highest common factor (HCF) of these numbers is 2.

The variable parts are $x^2$, $x^3$, and $x^4$. The lowest power of $x$ is $x^2$.

Therefore, the common factor for all terms is $2x^2$.

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Now, we factor out the common factor $2x^2$ from each term:

$10x^2 = 2x^2 \times 5$

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

$14x^4 = 2x^2 \times (7x^2)$

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Putting it together, we get:

$10x^2 - 18x^3 + 14x^4 = 2x^2(5 - 9x + 7x^2)$

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We can rearrange the terms inside the parenthesis in descending powers of $x$:

The factorised form is $2x^2(7x^2 - 9x + 5)$.

The given expression is $6xy - 4y + 6 - 9x$.

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This expression contains four terms. We can try to factorise it by grouping the terms.

Let's group the first two terms and the last two terms:

$(6xy - 4y) + (6 - 9x)$

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Factor out the common factor from the first group $(6xy - 4y)$. The common factor is $2y$:

$6xy - 4y = 2y(3x - 2)$

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Factor out the common factor from the second group $(6 - 9x)$. The common factor is $3$:

$6 - 9x = 3(2 - 3x)$

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Now substitute these back into the grouped expression:

$2y(3x - 2) + 3(2 - 3x)$

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Observe the terms inside the parentheses: $(3x - 2)$ and $(2 - 3x)$. They are negatives of each other, i.e., $(2 - 3x) = -(3x - 2)$.

Rewrite the expression by changing the sign of the second term and reversing the terms inside the parenthesis:

$2y(3x - 2) - 3(3x - 2)$

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Now we have a common binomial factor $(3x - 2)$ in both terms. Factor out $(3x - 2)$:

$(3x - 2)(2y - 3)$

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Alternatively, we could group the terms differently, for example, $(6xy - 9x) + (-4y + 6)$.

From $(6xy - 9x)$, the common factor is $3x$: $3x(2y - 3)$.

From $(-4y + 6)$, the common factor is $-2$: $-2(2y - 3)$.

So the expression becomes $3x(2y - 3) - 2(2y - 3)$.

Factoring out the common binomial factor $(2y - 3)$ gives $(2y - 3)(3x - 2)$.

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Both methods yield the same factorised form.

The factorised form is $(3x - 2)(2y - 3)$.

To find the common factors, we find the highest common factor (HCF) of the numerical coefficients and the lowest power of each variable that appears in all terms.

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(i) 12x, 36

Numerical coefficients: 12 and 36. HCF(12, 36) = 12.

Variables: 'x' is in the first term but not the second.

Common factor is the HCF of the coefficients and the variables common to all terms with their lowest power.

Common factor = 12.

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(ii) 2y, 22xy

Numerical coefficients: 2 and 22. HCF(2, 22) = 2.

Variables: 'y' is in both terms. Lowest power of y is $y^1 = y$. 'x' is only in the second term.

Common factor = numerical HCF $\times$ common variables with lowest power.

Common factor = $2 \times y$ = $2y$.

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(iii) 14 pq, 28p²q²

Numerical coefficients: 14 and 28. HCF(14, 28) = 14.

Variables: 'p' is in both terms. Lowest power of p is $p^1 = p$. 'q' is in both terms. Lowest power of q is $q^1 = q$.

Common factor = numerical HCF $\times$ common variables with lowest power.

Common factor = $14 \times p \times q$ = $14pq$.

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(iv) 2x, 3x², 4

Numerical coefficients: 2, 3, and 4. HCF(2, 3, 4) = 1.

Variables: 'x' is in the first and second terms, but not in the third term.

Since there are no variables common to all three terms, the common factor is just the HCF of the numerical coefficients.

Common factor = 1.

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(v) 6abc, 24ab², 12 a²b

Numerical coefficients: 6, 24, and 12. HCF(6, 24, 12) = 6.

Variables: 'a' is in all terms. Lowest power of a is $a^1 = a$. 'b' is in all terms. Lowest power of b is $b^1 = b$. 'c' is only in the first term.

Common factor = numerical HCF $\times$ common variables with lowest power.

Common factor = $6 \times a \times b$ = $6ab$.

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(vi) 16x³, – 4x², 32x

Numerical coefficients: 16, -4, and 32. The HCF of the absolute values (16, 4, 32) is 4.

Variables: 'x' is in all terms. The powers of x are 3, 2, and 1. The lowest power of x is $x^1 = x$.

Common factor = numerical HCF $\times$ common variables with lowest power.

Common factor = $4 \times x$ = $4x$.

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(vii) 10pq, 20qr, 30rp

Numerical coefficients: 10, 20, and 30. HCF(10, 20, 30) = 10.

Variables: 'p' is in the first and third terms. 'q' is in the first and second terms. 'r' is in the second and third terms.

There is no variable that appears in all three terms.

Common factor is just the HCF of the numerical coefficients.

Common factor = 10.

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(viii) 3x²y³, 10x³y², 6x²y²z

Numerical coefficients: 3, 10, and 6. HCF(3, 10, 6) = 1.

Variables: 'x' is in all terms. The powers of x are 2, 3, and 2. The lowest power of x is $x^2$. 'y' is in all terms. The powers of y are 3, 2, and 2. The lowest power of y is $y^2$. 'z' is only in the third term.

Common factor = numerical HCF $\times$ common variables with lowest power.

Common factor = $1 \times x^2 \times y^2$ = $x^2y^2$.

(i) $7x - 42$

The common factor of 7 and 42 is 7.

$7x - 42 = 7(x) - 7(6) = 7(x - 6)$

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(ii) $6p - 12q$

The common factor of 6 and 12 is 6.

$6p - 12q = 6(p) - 6(2q) = 6(p - 2q)$

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(iii) $7a^2 + 14a$

The common factor of 7 and 14 is 7. The common variable with the lowest power is $a^1 = a$.

The common factor is $7a$.

$7a^2 + 14a = 7a(a) + 7a(2) = 7a(a + 2)$

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(iv) $-16z + 20z^3$

The common factor of -16 and 20 is 4 (or -4). The common variable with the lowest power is $z^1 = z$.

Let's take the common factor as $4z$.

$-16z + 20z^3 = 4z(-4) + 4z(5z^2) = 4z(-4 + 5z^2)$

Alternatively, taking $-4z$ as the common factor:

$-16z + 20z^3 = -4z(4) + (-4z)(-5z^2) = -4z(4 - 5z^2)$

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(v) $20l^2m + 30 alm$

The common factor of 20 and 30 is 10.

The common variable $l$ has powers $l^2$ and $l^1$. Lowest power is $l^1 = l$.

The common variable $m$ has powers $m^1$ and $m^1$. Lowest power is $m^1 = m$.

The common factor is $10lm$.

$20l^2m + 30alm = 10lm(2l) + 10lm(3a) = 10lm(2l + 3a)$

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(vi) $5x^2y - 15xy^2$

The common factor of 5 and 15 is 5.

The common variable $x$ has powers $x^2$ and $x^1$. Lowest power is $x^1 = x$.

The common variable $y$ has powers $y^1$ and $y^2$. Lowest power is $y^1 = y$.

The common factor is $5xy$.

$5x^2y - 15xy^2 = 5xy(x) - 5xy(3y) = 5xy(x - 3y)$

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(vii) $10a^2 - 15b^2 + 20c^2$

The common factor of 10, 15, and 20 is 5.

The variables $a^2$, $b^2$, and $c^2$ are not common to all three terms.

The common factor is 5.

$10a^2 - 15b^2 + 20c^2 = 5(2a^2) - 5(3b^2) + 5(4c^2) = 5(2a^2 - 3b^2 + 4c^2)$

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(viii) $-4a^2 + 4ab - 4ca$

The common factor of -4, 4, and -4 is 4 (or -4).

The common variable is $a$ (it is present in all terms). The lowest power is $a^1 = a$.

The common factor is $4a$.

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

Alternatively, taking $-4a$ as the common factor:

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

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(ix) $x^2yz + xy^2z + xyz^2$

The numerical coefficient is 1 for all terms.

The common variable $x$ has powers 2, 1, 1. Lowest power is $x^1 = x$.

The common variable $y$ has powers 1, 2, 1. Lowest power is $y^1 = y$.

The common variable $z$ has powers 1, 1, 2. Lowest power is $z^1 = z$.

The common factor is $xyz$.

$x^2yz + xy^2z + xyz^2 = xyz(x) + xyz(y) + xyz(z) = xyz(x + y + z)$

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(x) $ax^2y + bxy^2 + cxyz$

The numerical coefficients are a, b, and c. Assuming they don't have a common numerical factor other than 1.

The common variable $x$ has powers 2, 1, 1. Lowest power is $x^1 = x$.

The common variable $y$ has powers 1, 2, 1. Lowest power is $y^1 = y$.

The variable $z$ is only in the third term.

The common factor is $xy$.

$ax^2y + bxy^2 + cxyz = xy(ax) + xy(by) + xy(cz) = xy(ax + by + cz)$

We will factorise the expressions by grouping terms and finding common factors.

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(i) $x^2 + xy + 8x + 8y$

Group the terms:

$(x^2 + xy) + (8x + 8y)$

Factor out the common factor from the first group ($x$) and the second group ($8$):

$x(x + y) + 8(x + y)$

Now, $(x + y)$ is a common binomial factor. Factor it out:

$(x + y)(x + 8)$

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(ii) $15xy – 6x + 5y – 2$

Group the terms:

$(15xy – 6x) + (5y – 2)$

Factor out the common factor from the first group ($3x$) and the second group ($1$):

$3x(5y - 2) + 1(5y - 2)$

Now, $(5y - 2)$ is a common binomial factor. Factor it out:

$(5y - 2)(3x + 1)$

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(iii) $ax + bx – ay – by$

Group the terms:

$(ax + bx) + (-ay – by)$

Factor out the common factor from the first group ($x$) and the second group ($-y$):

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

Now, $(a + b)$ is a common binomial factor. Factor it out:

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

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(iv) $15pq + 15 + 9q + 25p$

Rearrange the terms to group those with common factors:

$15pq + 9q + 25p + 15$

Group the terms:

$(15pq + 9q) + (25p + 15)$

Factor out the common factor from the first group ($3q$) and the second group ($5$):

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

Now, $(5p + 3)$ is a common binomial factor. Factor it out:

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

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(v) $z – 7 + 7xy – xyz$

Rearrange the terms to group those with common factors:

$z - xyz + 7xy - 7$

Group the terms:

$(z - xyz) + (7xy - 7)$

Factor out the common factor from the first group ($z$) and the second group ($7$):

$z(1 - xy) + 7(xy - 1)$

Note that $(xy - 1) = -(1 - xy)$. Rewrite the expression:

$z(1 - xy) - 7(1 - xy)$

Now, $(1 - xy)$ is a common binomial factor. Factor it out:

$(1 - xy)(z - 7)$

The given expression is $x^2 + 8x + 16$. This is a quadratic trinomial.

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Method 1: By Splitting the Middle Term

We look for two numbers whose product is equal to the constant term (16) and whose sum is equal to the coefficient of the middle term (8).

Let the two numbers be $p$ and $q$. We need $p \times q = 16$ and $p + q = 8$.

By checking the factors of 16 (1, 2, 4, 8, 16), we find that $4 \times 4 = 16$ and $4 + 4 = 8$.

So the two numbers are 4 and 4.

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Now, we split the middle term ($8x$) using these numbers:

$x^2 + 8x + 16 = x^2 + 4x + 4x + 16$

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Group the terms and factor by grouping:

$(x^2 + 4x) + (4x + 16)$

Factor out the common factor from each group:

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

Now, $(x + 4)$ is a common binomial factor. Factor it out:

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

Which can be written as:

$(x + 4)^2$

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Method 2: Using Algebraic Identity

Observe the expression $x^2 + 8x + 16$. It looks like the form of a perfect square trinomial, which is $a^2 + 2ab + b^2 = (a + b)^2$.

Compare the given expression with the identity:

First term: $x^2$ corresponds to $a^2$. This suggests $a = x$.

Last term: $16$ corresponds to $b^2$. Since $16 = 4^2$, this suggests $b = 4$ (assuming $b$ is positive).

Now check the middle term: The middle term in the identity is $2ab$.

Substitute the values of $a$ and $b$ we found: $2ab = 2(x)(4) = 8x$.

This matches the middle term in the given expression ($+8x$).

Since the expression is in the form $a^2 + 2ab + b^2$ with $a=x$ and $b=4$, it can be factorised as $(a + b)^2$.

$x^2 + 8x + 16 = (x + 4)^2$

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Both methods give the same result.

The factorised form is $(x + 4)^2$.

The given expression is $4y^2 - 12y + 9$. This is a quadratic trinomial.

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Method 1: By Splitting the Middle Term

We look for two numbers whose product is the product of the coefficient of $y^2$ (4) and the constant term (9), i.e., $4 \times 9 = 36$. Their sum must be the coefficient of the middle term (-12).

Let the two numbers be $p$ and $q$. We need $p \times q = 36$ and $p + q = -12$.

Since the product is positive and the sum is negative, both numbers must be negative. The pairs of negative factors of 36 are (-1, -36), (-2, -18), (-3, -12), (-4, -9), (-6, -6).

The pair $(-6, -6)$ satisfies the conditions: $(-6) \times (-6) = 36$ and $(-6) + (-6) = -12$.

So the two numbers are -6 and -6.

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Now, we split the middle term ($-12y$) using these numbers:

$4y^2 - 12y + 9 = 4y^2 - 6y - 6y + 9$

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Group the terms and factor by grouping:

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

Factor out the common factor from each group:

From the first group, the common factor is $2y$: $2y(2y - 3)$.

From the second group, the common factor is $-3$: $-3(2y - 3)$.

The expression becomes:

$2y(2y - 3) - 3(2y - 3)$

Now, $(2y - 3)$ is a common binomial factor. Factor it out:

$(2y - 3)(2y - 3)$

Which can be written as:

$(2y - 3)^2$

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Method 2: Using Algebraic Identity

Observe the expression $4y^2 - 12y + 9$. It looks like the form of a perfect square trinomial, which is $a^2 - 2ab + b^2 = (a - b)^2$.

Compare the given expression with the identity:

First term: $4y^2$ corresponds to $a^2$. Since $4y^2 = (2y)^2$, this suggests $a = 2y$.

Last term: $9$ corresponds to $b^2$. Since $9 = 3^2$, this suggests $b = 3$ (assuming $b$ is positive).

Now check the middle term: The middle term in the identity is $-2ab$.

Substitute the values of $a$ and $b$ we found: $-2ab = -2(2y)(3) = -12y$.

This matches the middle term in the given expression ($-12y$).

Since the expression is in the form $a^2 - 2ab + b^2$ with $a=2y$ and $b=3$, it can be factorised as $(a - b)^2$.

$4y^2 - 12y + 9 = (2y - 3)^2$

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Both methods give the same result.

The factorised form is $(2y - 3)^2$.

The given expression is $49p^2 - 36$.

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We observe that this expression is in the form of a difference of two squares, which can be factorised using the algebraic identity: $a^2 - b^2 = (a - b)(a + b)$.

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We need to express each term in the form of a square:

The first term is $49p^2$. We can write this as $(7p)^2$, since $(7p)^2 = 7^2 \times p^2 = 49p^2$.

So, we can consider $a = 7p$.

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The second term is $36$. We can write this as $6^2$, since $6^2 = 36$.

So, we can consider $b = 6$.

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Now, substitute $a = 7p$ and $b = 6$ into the identity $a^2 - b^2 = (a - b)(a + b)$.

$49p^2 - 36 = (7p)^2 - 6^2 = (7p - 6)(7p + 6)$

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The factorised form is $(7p - 6)(7p + 6)$.

The given expression is $a^2 - 2ab + b^2 - c^2$.

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Observe the first three terms: $a^2 - 2ab + b^2$. This is a perfect square trinomial, which fits the algebraic identity $a^2 - 2ab + b^2 = (a - b)^2$.

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Rewrite the given expression by replacing the first three terms with their factorised form:

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

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Now the expression is in the form of a difference of two squares: $(a - b)^2 - c^2$.

This matches the algebraic identity $x^2 - y^2 = (x - y)(x + y)$.

Here, we can consider $x = (a - b)$ and $y = c$.

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Substitute these into the identity $x^2 - y^2 = (x - y)(x + y)$:

$(a - b)^2 - c^2 = \big((a - b) - c\big)\big((a - b) + c\big)$

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Simplify the terms inside the parentheses:

$(a - b - c)(a - b + c)$

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The factorised form is $(a - b - c)(a - b + c)$.

The given expression is $m^4 - 256$.

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This expression is a difference of two terms, and both terms are perfect squares.

We can write $m^4$ as $(m^2)^2$.

We can write 256 as $16^2$ (since $16 \times 16 = 256$).

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So the expression can be written as $(m^2)^2 - 16^2$.

This is in the form of a difference of two squares, $a^2 - b^2$, where $a = m^2$ and $b = 16$.

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

$(m^2)^2 - 16^2 = (m^2 - 16)(m^2 + 16)$

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Now, look at the factor $(m^2 - 16)$. This is also a difference of two squares.

We can write $m^2$ as $(m)^2$.

We can write 16 as $4^2$.

So $(m^2 - 16)$ can be written as $(m)^2 - 4^2$.

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Using the identity $a^2 - b^2 = (a - b)(a + b)$ again, with $a = m$ and $b = 4$, we factorise $(m^2 - 16)$:

$m^2 - 16 = (m - 4)(m + 4)$

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The factor $(m^2 + 16)$ cannot be factorised further into real linear factors.

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Substitute the factorised form of $(m^2 - 16)$ back into the expression:

$m^4 - 256 = (m^2 - 16)(m^2 + 16) = (m - 4)(m + 4)(m^2 + 16)$

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The factorised form is $(m - 4)(m + 4)(m^2 + 16)$.

The given expression is $x^2 + 5x + 6$. This is a quadratic trinomial of the form $x^2 + bx + c$, where $b=5$ and $c=6$.

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To factorise this, we need to find two numbers that multiply to give the constant term (6) and add up to give the coefficient of the middle term (5).

Let the two numbers be $p$ and $q$. We are looking for $p$ and $q$ such that:

$p \times q = 6$

$p + q = 5$

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Let's consider the pairs of factors of 6:

Factors of 6 are (1, 6), (2, 3), (-1, -6), (-2, -3).

Now let's check their sums:

$1 + 6 = 7$ (Not 5)

$2 + 3 = 5$ (This is the correct pair)

$-1 + (-6) = -7$ (Not 5)

$-2 + (-3) = -5$ (Not 5)

The two numbers we are looking for are 2 and 3.

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We can now write the trinomial as a product of two binomials using these numbers:

$x^2 + 5x + 6 = (x + p)(x + q)$

Substitute $p=2$ and $q=3$:

$x^2 + 5x + 6 = (x + 2)(x + 3)$

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Alternatively, we can use the method of splitting the middle term. Split $5x$ into $2x + 3x$ based on the numbers found above:

$x^2 + 5x + 6 = x^2 + 2x + 3x + 6$

Group the terms:

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

Factor out the common factor from each group:

$x(x + 2) + 3(x + 2)$

Factor out the common binomial factor $(x + 2)$:

$(x + 2)(x + 3)$

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Both methods give the same result.

The factorised form is $(x + 2)(x + 3)$.

The given expression is $y^2 - 7y + 12$. This is a quadratic trinomial of the form $y^2 + by + c$, where $b=-7$ and $c=12$.

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To factorise this, we need to find two numbers that multiply to give the constant term (12) and add up to give the coefficient of the middle term (-7).

Let the two numbers be $p$ and $q$. We are looking for $p$ and $q$ such that:

$p \times q = 12$

$p + q = -7$

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Let's consider the pairs of integer factors of 12. Since the product is positive (12) and the sum is negative (-7), both numbers must be negative.

Pairs of negative factors of 12:

$(-1, -12) \implies -1 + (-12) = -13$

$(-2, -6) \implies -2 + (-6) = -8$

$(-3, -4) \implies -3 + (-4) = -7$

The pair $(-3, -4)$ satisfies both conditions.

So the two numbers we are looking for are -3 and -4.

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Method 1: Direct factorisation

The trinomial $y^2 + by + c$ can be factorised as $(y + p)(y + q)$ where $p$ and $q$ are the numbers found above.

$y^2 - 7y + 12 = (y + (-3))(y + (-4)) = (y - 3)(y - 4)$

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Method 2: Splitting the middle term

Split the middle term $-7y$ into $-3y$ and $-4y$:

$y^2 - 7y + 12 = y^2 - 3y - 4y + 12$

Group the terms:

$(y^2 - 3y) + (-4y + 12)$

Factor out the common factor from each group:

$y(y - 3) - 4(y - 3)$

Factor out the common binomial factor $(y - 3)$:

$(y - 3)(y - 4)$

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Both methods yield the same result.

The factorised form is $(y - 3)(y - 4)$.

The given expression is $z^2 - 4z - 12$. This is a quadratic trinomial of the form $z^2 + bz + c$, where $b = -4$ and $c = -12$.

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To factorise this expression, we need to find two numbers, let's call them $p$ and $q$, such that their product $p \times q$ is equal to the constant term ($-12$) and their sum $p + q$ is equal to the coefficient of the middle term ($-4$).

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

$p \times q = -12$

$p + q = -4$

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Let's list the pairs of integer factors of -12. Since the product is negative, one factor must be positive and the other must be negative.

Pairs of factors $(p, q)$ of -12 and their sums:

$(1, -12) \implies 1 + (-12) = -11$ (Not -4)

$(-1, 12) \implies -1 + 12 = 11$ (Not -4)

$(2, -6) \implies 2 + (-6) = -4$ (This is the correct pair)

$(-2, 6) \implies -2 + 6 = 4$ (Not -4)

$(3, -4) \implies 3 + (-4) = -1$ (Not -4)

$(-3, 4) \implies -3 + 4 = 1$ (Not -4)

The two numbers that satisfy the conditions are 2 and -6.

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Method 1: Using the numbers directly

For a trinomial of the form $z^2 + bz + c$, if we find two numbers $p$ and $q$ such that $p \times q = c$ and $p + q = b$, then the factorisation is $(z + p)(z + q)$.

Using $p=2$ and $q=-6$:

$z^2 - 4z - 12 = (z + 2)(z + (-6)) = (z + 2)(z - 6)$

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Method 2: Splitting the middle term

We split the middle term $-4z$ into $2z$ and $-6z$ based on the numbers we found.

$z^2 - 4z - 12 = z^2 + 2z - 6z - 12$

Now, group the terms:

$(z^2 + 2z) + (-6z - 12)$

Factor out the common factor from the first group ($z$) and the second group ($-6$):

$z(z + 2) - 6(z + 2)$

Now, $(z + 2)$ is a common binomial factor. Factor it out:

$(z + 2)(z - 6)$

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Both methods yield the same result.

The factorised form is $(z + 2)(z - 6)$.

The given expression is $3m^2 + 9m + 6$. This is a quadratic trinomial.

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First, we look for a common factor among all the terms (3, 9, and 6).

The highest common factor (HCF) of 3, 9, and 6 is 3.

Factor out the common factor 3 from the expression:

$3m^2 + 9m + 6 = 3(m^2 + 3m + 2)$

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Now, we need to factorise the quadratic trinomial inside the parenthesis: $m^2 + 3m + 2$.

This is in the form $m^2 + bm + c$, where $b=3$ and $c=2$.

We need to find two numbers that multiply to the constant term (2) and add up to the coefficient of the middle term (3).

Let the two numbers be $p$ and $q$. We are looking for $p$ and $q$ such that:

$p \times q = 2$

$p + q = 3$

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Consider the integer pairs of factors of 2:

(1, 2)

(-1, -2)

Check their sums:

$1 + 2 = 3$ (This pair works)

$-1 + (-2) = -3$ (This pair does not work)

The two numbers are 1 and 2.

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Now, we factorise the trinomial $m^2 + 3m + 2$ using these numbers. The factors will be of the form $(m + p)(m + q)$.

$m^2 + 3m + 2 = (m + 1)(m + 2)$

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Finally, substitute this back into the original expression, including the common factor 3:

$3m^2 + 9m + 6 = 3(m^2 + 3m + 2) = 3(m + 1)(m + 2)$

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The factorised form is $3(m + 1)(m + 2)$.

(i) $a^2 + 8a + 16$

This expression is in the form $a^2 + 2ab + b^2$.

Compare with $a^2 + 2ab + b^2$: $(a)^2 + 2(a)(4) + (4)^2$.

This matches the identity $(a + b)^2$ with $a=a$ and $b=4$.

$a^2 + 8a + 16 = (a + 4)^2$

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(ii) $p^2 – 10p + 25$

This expression is in the form $a^2 - 2ab + b^2$.

Compare with $a^2 - 2ab + b^2$: $(p)^2 - 2(p)(5) + (5)^2$.

This matches the identity $(a - b)^2$ with $a=p$ and $b=5$.

$p^2 – 10p + 25 = (p - 5)^2$

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(iii) $25m^2 + 30m + 9$

This expression is in the form $a^2 + 2ab + b^2$.

Compare with $a^2 + 2ab + b^2$: $(5m)^2 + 2(5m)(3) + (3)^2$.

This matches the identity $(a + b)^2$ with $a=5m$ and $b=3$.

$25m^2 + 30m + 9 = (5m + 3)^2$

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(iv) $49y^2 + 84yz + 36z^2$

This expression is in the form $a^2 + 2ab + b^2$.

Compare with $a^2 + 2ab + b^2$: $(7y)^2 + 2(7y)(6z) + (6z)^2$.

This matches the identity $(a + b)^2$ with $a=7y$ and $b=6z$.

$49y^2 + 84yz + 36z^2 = (7y + 6z)^2$

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(v) $4x^2 – 8x + 4$

First, factor out the common factor 4:

$4x^2 – 8x + 4 = 4(x^2 - 2x + 1)$

Now, factorise the trinomial $x^2 - 2x + 1$. This is in the form $a^2 - 2ab + b^2$.

Compare with $a^2 - 2ab + b^2$: $(x)^2 - 2(x)(1) + (1)^2$.

This matches the identity $(a - b)^2$ with $a=x$ and $b=1$.

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

Substitute this back into the expression:

$4(x^2 - 2x + 1) = 4(x - 1)^2$

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(vi) $121b^2 – 88bc + 16c^2$

This expression is in the form $a^2 - 2ab + b^2$.

Compare with $a^2 - 2ab + b^2$: $(11b)^2 - 2(11b)(4c) + (4c)^2$.

This matches the identity $(a - b)^2$ with $a=11b$ and $b=4c$.

$121b^2 – 88bc + 16c^2 = (11b - 4c)^2$

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(vii) $(l + m)^2 – 4lm$

Expand the first term $(l + m)^2$ using the identity $(a + b)^2 = a^2 + 2ab + b^2$:

$(l + m)^2 = l^2 + 2lm + m^2$

Substitute this back into the expression:

$(l^2 + 2lm + m^2) - 4lm$

Combine the like terms ($2lm - 4lm$):

$l^2 - 2lm + m^2$

Now, factorise the resulting expression. This is in the form $a^2 - 2ab + b^2$.

Compare with $a^2 - 2ab + b^2$: $(l)^2 - 2(l)(m) + (m)^2$.

This matches the identity $(a - b)^2$ with $a=l$ and $b=m$.

$l^2 - 2lm + m^2 = (l - m)^2$

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(viii) $a^4 + 2a^2b^2 + b^4$

This expression can be seen as a perfect square trinomial.

Let $x = a^2$ and $y = b^2$. Then the expression becomes $x^2 + 2xy + y^2$.

This matches the identity $x^2 + 2xy + y^2 = (x + y)^2$.

Substitute back $x = a^2$ and $y = b^2$:

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

(i) $4p^2 – 9q^2$

This is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.

Here $a^2 = 4p^2 = (2p)^2$, so $a = 2p$.

And $b^2 = 9q^2 = (3q)^2$, so $b = 3q$.

Therefore, $4p^2 – 9q^2 = (2p - 3q)(2p + 3q)$.

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(ii) $63a^2 – 112b^2$

First, find the highest common factor of the numerical coefficients 63 and 112.

Factors of 63: 1, 3, 7, 9, 21, 63.

Factors of 112: 1, 2, 4, 7, 8, 14, 16, 28, 56, 112.

The HCF is 7.

Factor out 7:

$63a^2 – 112b^2 = 7(9a^2 - 16b^2)$

Now, the expression in the parenthesis is a difference of squares.

$9a^2 = (3a)^2$ and $16b^2 = (4b)^2$.

So, $9a^2 - 16b^2 = (3a - 4b)(3a + 4b)$.

Therefore, $63a^2 – 112b^2 = 7(3a - 4b)(3a + 4b)$.

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(iii) $49x^2 – 36$

This is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.

Here $a^2 = 49x^2 = (7x)^2$, so $a = 7x$.

And $b^2 = 36 = 6^2$, so $b = 6$.

Therefore, $49x^2 – 36 = (7x - 6)(7x + 6)$.

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(iv) $16x^5 – 144x^3$

First, find the highest common factor of the terms. The numerical coefficients are 16 and 144, HCF(16, 144) = 16. The variable parts are $x^5$ and $x^3$, lowest power is $x^3$.

The common factor is $16x^3$.

Factor out $16x^3$:

$16x^5 – 144x^3 = 16x^3(x^2 - 9)$

Now, the expression in the parenthesis is a difference of squares.

$x^2 = (x)^2$ and $9 = 3^2$.

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

Therefore, $16x^5 – 144x^3 = 16x^3(x - 3)(x + 3)$.

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(v) $(l + m)^2 – (l – m)^2$

This is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.

Here $a = (l + m)$ and $b = (l – m)$.

$a - b = (l + m) - (l - m) = l + m - l + m = 2m$

$a + b = (l + m) + (l - m) = l + m + l - m = 2l$

Therefore, $(l + m)^2 – (l – m)^2 = (2m)(2l) = 4lm$.

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(vi) $9x^2y^2 – 16$

This is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.

Here $a^2 = 9x^2y^2 = (3xy)^2$, so $a = 3xy$.

And $b^2 = 16 = 4^2$, so $b = 4$.

Therefore, $9x^2y^2 – 16 = (3xy - 4)(3xy + 4)$.

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(vii) $(x^2 – 2xy + y^2) – z^2$

Observe the first three terms: $x^2 – 2xy + y^2$. This is a perfect square trinomial, $(x - y)^2$.

Replace the trinomial with its factorised form:

$(x^2 – 2xy + y^2) – z^2 = (x - y)^2 - z^2$

Now, this is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.

Here $a = (x - y)$ and $b = z$.

Therefore, $(x - y)^2 - z^2 = \big((x - y) - z\big)\big((x - y) + z\big) = (x - y - z)(x - y + z)$.

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(viii) $25a^2 – 4b^2 + 28bc – 49c^2$

Rearrange the terms to group the terms involving b and c:

$25a^2 - (4b^2 - 28bc + 49c^2)$

Observe the expression in the parenthesis: $4b^2 - 28bc + 49c^2$. This is a perfect square trinomial.

Compare with $A^2 - 2AB + B^2 = (A - B)^2$.

$A^2 = 4b^2 = (2b)^2$, so $A = 2b$.

$B^2 = 49c^2 = (7c)^2$, so $B = 7c$.

Check the middle term: $2AB = 2(2b)(7c) = 28bc$. This matches.

So, $4b^2 - 28bc + 49c^2 = (2b - 7c)^2$.

Substitute this back into the main expression:

$25a^2 - (2b - 7c)^2$

Now, this is in the form of a difference of two squares, $X^2 - Y^2 = (X - Y)(X + Y)$.

Here $X^2 = 25a^2 = (5a)^2$, so $X = 5a$.

And $Y^2 = (2b - 7c)^2$, so $Y = (2b - 7c)$.

Apply the identity:

$25a^2 - (2b - 7c)^2 = \big(5a - (2b - 7c)\big)\big(5a + (2b - 7c)\big)$

Simplify the terms inside the brackets:

$\big(5a - 2b + 7c\big)\big(5a + 2b - 7c\big)$

The factorised form is $(5a - 2b + 7c)(5a + 2b - 7c)$.

(i) $ax^2 + bx$

The common factor in both terms is $x$.

$ax^2 + bx = x(ax) + x(b) = x(ax + b)$

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(ii) $7p^2 + 21q^2$

The common numerical factor of 7 and 21 is 7.

There is no common variable.

$7p^2 + 21q^2 = 7(p^2) + 7(3q^2) = 7(p^2 + 3q^2)$

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(iii) $2x^3 + 2xy^2 + 2xz^2$

The common factor in all terms is $2x$.

$2x^3 + 2xy^2 + 2xz^2 = 2x(x^2) + 2x(y^2) + 2x(z^2) = 2x(x^2 + y^2 + z^2)$

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(iv) $am^2 + bm^2 + bn^2 + an^2$

Group the terms:

$(am^2 + bm^2) + (bn^2 + an^2)$

Factor out common factors from each group:

$m^2(a + b) + n^2(b + a)$

Since $(b + a) = (a + b)$, we have a common binomial factor $(a + b)$.

$m^2(a + b) + n^2(a + b) = (a + b)(m^2 + n^2)$

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(v) $(lm + l) + m + 1$

The terms are already grouped partially.

Factor out the common factor $l$ from the first group $(lm + l)$.

$l(m + 1) + (m + 1)$

Note that the second group $(m + 1)$ can be written as $1(m + 1)$.

$l(m + 1) + 1(m + 1)$

Now, $(m + 1)$ is a common binomial factor. Factor it out:

$(m + 1)(l + 1)$

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(vi) $y (y + z) + 9 (y + z)$

The expression already has a common binomial factor $(y + z)$.

Factor out $(y + z)$:

$(y + z)(y + 9)$

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(vii) $5y^2 – 20y – 8z + 2yz$

Group the terms:

$(5y^2 – 20y) + (-8z + 2yz)$

Factor out common factors from each group:

$5y(y - 4) + 2z(-4 + y)$

Since $(-4 + y) = (y - 4)$, we have a common binomial factor $(y - 4)$.

$5y(y - 4) + 2z(y - 4) = (y - 4)(5y + 2z)$

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(viii) $10ab + 4a + 5b + 2$

Group the terms:

$(10ab + 4a) + (5b + 2)$

Factor out common factors from each group:

$2a(5b + 2) + 1(5b + 2)$

Now, $(5b + 2)$ is a common binomial factor. Factor it out:

$(5b + 2)(2a + 1)$

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(ix) $6xy – 4y + 6 – 9x$

Group the terms:

$(6xy – 4y) + (6 – 9x)$

Factor out common factors from each group:

$2y(3x - 2) + 3(2 - 3x)$

Note that $(2 - 3x) = -(3x - 2)$. Rewrite the second term:

$2y(3x - 2) - 3(3x - 2)$

Now, $(3x - 2)$ is a common binomial factor. Factor it out:

$(3x - 2)(2y - 3)$

(i) $a^4 – b^4$

This expression can be written as a difference of squares: $(a^2)^2 - (b^2)^2$.

Using the identity $x^2 - y^2 = (x - y)(x + y)$ with $x=a^2$ and $y=b^2$:

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

Now, the factor $(a^2 - b^2)$ is also a difference of squares: $(a)^2 - (b)^2$.

Using the identity again with $x=a$ and $y=b$:

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

The factor $(a^2 + b^2)$ cannot be factorised further into real linear factors.

Substitute the factorisation of $(a^2 - b^2)$ back into the expression:

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

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(ii) $p^4 – 81$

This expression can be written as a difference of squares: $(p^2)^2 - (9)^2$.

Using the identity $x^2 - y^2 = (x - y)(x + y)$ with $x=p^2$ and $y=9$:

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

Now, the factor $(p^2 - 9)$ is also a difference of squares: $(p)^2 - (3)^2$.

Using the identity again with $x=p$ and $y=3$:

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

The factor $(p^2 + 9)$ cannot be factorised further into real linear factors.

Substitute the factorisation of $(p^2 - 9)$ back into the expression:

$p^4 – 81 = (p - 3)(p + 3)(p^2 + 9)$

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(iii) $x^4 – (y + z)^4$

This expression can be written as a difference of squares: $(x^2)^2 - \big((y + z)^2\big)^2$.

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=x^2$ and $b=(y + z)^2$:

$(x^2)^2 - \big((y + z)^2\big)^2 = \big(x^2 - (y + z)^2\big)\big(x^2 + (y + z)^2\big)$

Now, the factor $\big(x^2 - (y + z)^2\big)$ is a difference of squares: $(x)^2 - (y + z)^2$.

Using the identity again with $a=x$ and $b=(y + z)$:

$x^2 - (y + z)^2 = \big(x - (y + z)\big)\big(x + (y + z)\big) = (x - y - z)(x + y + z)$

Consider the second factor $\big(x^2 + (y + z)^2\big)$. Expand $(y+z)^2 = y^2 + 2yz + z^2$.

$\big(x^2 + (y + z)^2\big) = x^2 + y^2 + 2yz + z^2$. This cannot be easily factorised further.

Substitute the factorisation of $\big(x^2 - (y + z)^2\big)$ back into the expression:

$x^4 – (y + z)^4 = (x - y - z)(x + y + z)\big(x^2 + (y + z)^2\big)$

Which is equal to $(x - y - z)(x + y + z)(x^2 + y^2 + 2yz + z^2)$.

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(iv) $x^4 – (x – z)^4$

This expression can be written as a difference of squares: $(x^2)^2 - \big((x – z)^2\big)^2$.

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=x^2$ and $b=(x – z)^2$:

$(x^2)^2 - \big((x – z)^2\big)^2 = \big(x^2 - (x – z)^2\big)\big(x^2 + (x – z)^2\big)$

Now, the factor $\big(x^2 - (x – z)^2\big)$ is a difference of squares: $(x)^2 - (x - z)^2$.

Using the identity again with $a=x$ and $b=(x – z)$:

$x^2 - (x – z)^2 = \big(x - (x – z)\big)\big(x + (x – z)\big)$

Simplify the terms inside the parentheses:

$\big(x - x + z\big)\big(x + x - z\big) = (z)(2x - z)$

Consider the second factor $\big(x^2 + (x – z)^2\big)$. Expand $(x-z)^2 = x^2 - 2xz + z^2$.

$\big(x^2 + (x – z)^2\big) = x^2 + (x^2 - 2xz + z^2) = x^2 + x^2 - 2xz + z^2 = 2x^2 - 2xz + z^2$.

Substitute the factorisations back into the expression:

$x^4 – (x – z)^4 = (z)(2x - z)(2x^2 - 2xz + z^2)$

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(v) $a^4 – 2a^2b^2 + b^4$

This expression can be written as $(a^2)^2 - 2(a^2)(b^2) + (b^2)^2$.

This is in the form of a perfect square trinomial $X^2 - 2XY + Y^2 = (X - Y)^2$, where $X = a^2$ and $Y = b^2$.

Therefore, $(a^2)^2 - 2(a^2)(b^2) + (b^2)^2 = (a^2 - b^2)^2$.

We know that $a^2 - b^2 = (a - b)(a + b)$.

So, $(a^2 - b^2)^2 = \big((a - b)(a + b)\big)^2 = (a - b)^2 (a + b)^2$.

(i) $p^2 + 6p + 8$

We need to find two numbers whose product is 8 and whose sum is 6.

Let the numbers be 2 and 4.

$2 \times 4 = 8$

$2 + 4 = 6$

So, we can split the middle term $6p$ into $2p + 4p$, or directly write the factors.

Using the direct method:

$p^2 + 6p + 8 = (p + 2)(p + 4)$

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(ii) $q^2 – 10q + 21$

We need to find two numbers whose product is 21 and whose sum is -10.

Since the product is positive (21) and the sum is negative (-10), both numbers must be negative.

Let the numbers be -3 and -7.

$(-3) \times (-7) = 21$

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

So, we can split the middle term $-10q$ into $-3q - 7q$, or directly write the factors.

Using the direct method:

$q^2 – 10q + 21 = (q - 3)(q - 7)$

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(iii) $p^2 + 6p – 16$

We need to find two numbers whose product is -16 and whose sum is 6.

Since the product is negative (-16), one number must be positive and the other must be negative.

Let the numbers be 8 and -2.

$8 \times (-2) = -16$

$8 + (-2) = 6$

So, we can split the middle term $6p$ into $8p - 2p$, or directly write the factors.

Using the direct method:

$p^2 + 6p – 16 = (p + 8)(p - 2)$

(i) –20x⁴ ÷ 10x²

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We need to divide the monomial $-20x^4$ by the monomial $10x^2$.

This can be written as a fraction:

$\frac{-20x^4}{10x^2}$

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Divide the numerical coefficients:

$\frac{-20}{10} = -2$

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Divide the variable parts using the rule $a^m \div a^n = a^{m-n}$:

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

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Multiply the results from the numerical and variable divisions:

$-2 \times x^2 = -2x^2$

So, $-20x^4 \div 10x^2 = -2x^2$.

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(ii) 7x²y²z² ÷ 14xyz

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We need to divide the monomial $7x^2y^2z^2$ by the monomial $14xyz$.

This can be written as a fraction:

$\frac{7x^2y^2z^2}{14xyz}$

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Divide the numerical coefficients:

$\frac{7}{14} = \frac{1}{2}$

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Divide the variable parts for each variable separately using the rule $a^m \div a^n = a^{m-n}$:

$x^2 \div x = x^{2-1} = x^1 = x$

$y^2 \div y = y^{2-1} = y^1 = y$

$z^2 \div z = z^{2-1} = z^1 = z$

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Multiply the results from the numerical and variable divisions:

$\frac{1}{2} \times x \times y \times z = \frac{1}{2}xyz$

So, $7x^2y^2z^2 \div 14xyz = \frac{1}{2}xyz$.

The given division is $24(x^2yz + xy^2z + xyz^2) \div 8xyz$.

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Method 1: Factorising the numerator

First, factorise the expression inside the parenthesis in the numerator: $x^2yz + xy^2z + xyz^2$.

The common factor among $x^2yz$, $xy^2z$, and $xyz^2$ is $xyz$.

$x^2yz + xy^2z + xyz^2 = xyz(x) + xyz(y) + xyz(z) = xyz(x + y + z)$.

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Now, substitute this factorised form back into the numerator:

Numerator = $24(xyz(x + y + z)) = 24xyz(x + y + z)$.

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The division problem becomes:

$\frac{24xyz(x + y + z)}{8xyz}$

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Cancel the common factors in the numerator and the denominator. Both 24 and 8 are divisible by 8, and $xyz$ is common to both.

$\frac{\cancel{24}^{\;3} \cancel{xyz} (x + y + z)}{\cancel{8}_{\;1} \cancel{xyz}}$

The result is $3(x + y + z)$.

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Method 2: Dividing each term

First, distribute 24 into the parenthesis in the numerator:

$24(x^2yz + xy^2z + xyz^2) = 24x^2yz + 24xy^2z + 24xyz^2$

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Now, divide each term of this expression by the divisor $8xyz$:

$\frac{24x^2yz + 24xy^2z + 24xyz^2}{8xyz} = \frac{24x^2yz}{8xyz} + \frac{24xy^2z}{8xyz} + \frac{24xyz^2}{8xyz}$

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Perform each division separately:

$\frac{24x^2yz}{8xyz} = \frac{24}{8} \times \frac{x^2}{x} \times \frac{y}{y} \times \frac{z}{z} = 3 \times x^{2-1} \times y^{1-1} \times z^{1-1} = 3 \times x \times y^0 \times z^0 = 3x \times 1 \times 1 = 3x$

$\frac{24xy^2z}{8xyz} = \frac{24}{8} \times \frac{x}{x} \times \frac{y^2}{y} \times \frac{z}{z} = 3 \times x^{1-1} \times y^{2-1} \times z^{1-1} = 3 \times x^0 \times y \times z^0 = 3 \times 1 \times y \times 1 = 3y$

$\frac{24xyz^2}{8xyz} = \frac{24}{8} \times \frac{x}{x} \times \frac{y}{y} \times \frac{z^2}{z} = 3 \times x^{1-1} \times y^{1-1} \times z^{2-1} = 3 \times x^0 \times y^0 \times z = 3 \times 1 \times 1 \times z = 3z$

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Add the results of the individual divisions:

$3x + 3y + 3z$

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This expression can be factorised by taking out the common factor 3:

$3(x + y + z)$

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Both methods give the same result.

The result of the division is $3(x + y + z)$.

The given division is $44(x^4 – 5x^3 – 24x^2) \div 11x (x – 8)$.

We can write this as a fraction:

$\frac{44(x^4 – 5x^3 – 24x^2)}{11x (x – 8)}$

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First, let's factorise the expression in the numerator, specifically the part inside the parenthesis: $x^4 – 5x^3 – 24x^2$.

Observe that $x^2$ is a common factor in all terms:

$x^4 – 5x^3 – 24x^2 = x^2(x^2 – 5x – 24)$

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Now, we factorise the quadratic trinomial $x^2 – 5x – 24$. We look for two numbers that multiply to -24 and add up to -5.

The pairs of integer factors of -24 are (1, -24), (-1, 24), (2, -12), (-2, 12), (3, -8), (-3, 8), (4, -6), (-4, 6).

Let's check their sums:

$1 + (-24) = -23$

$(-1) + 24 = 23$

$2 + (-12) = -10$

$(-2) + 12 = 10$

$3 + (-8) = -5$

The numbers are 3 and -8. Their product is $3 \times (-8) = -24$ and their sum is $3 + (-8) = -5$.

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So, the trinomial factorises as:

$x^2 – 5x – 24 = (x + 3)(x - 8)$

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Substitute this back into the factored numerator:

$x^4 – 5x^3 – 24x^2 = x^2(x + 3)(x - 8)$

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Now the original numerator is $44 \times x^2(x + 3)(x - 8)$.

The division becomes:

$\frac{44x^2(x + 3)(x - 8)}{11x(x - 8)}$

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Cancel the common factors in the numerator and the denominator. Assume $x \neq 0$ and $x \neq 8$ so that the denominator is not zero.

Cancel the numerical factors: $\frac{\cancel{44}^{\;4}}{\cancel{11}_{\;1}} = 4$.

Cancel the variable factors: $\frac{\cancel{x^2}^{\;x}}{\cancel{x}_{\;1}} = x$.

Cancel the binomial factor: $\frac{\cancel{(x - 8)}}{\cancel{(x - 8)}} = 1$.

The remaining factors are 4, $x$, and $(x + 3)$.

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Multiply the remaining factors:

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

The result of the division is $4x(x + 3)$.

The given division is $z(5z^2 – 80) \div 5z(z + 4)$.

We can write this as a fraction:

$\frac{z(5z^2 - 80)}{5z(z + 4)}$

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First, we need to factorise the expression in the numerator, specifically the term inside the parenthesis: $5z^2 - 80$.

Find the common factor of 5 and 80, which is 5.

$5z^2 - 80 = 5(z^2 - 16)$

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Now, the expression inside the parenthesis, $z^2 - 16$, is a difference of squares. We can write it as $z^2 - 4^2$.

Using the identity $a^2 - b^2 = (a - b)(a + b)$, where $a=z$ and $b=4$:

$z^2 - 16 = (z - 4)(z + 4)$

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Substitute this factorisation back into the expression $5(z^2 - 16)$:

$5z^2 - 80 = 5(z - 4)(z + 4)$

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Now, substitute this fully factorised form back into the numerator of the original fraction:

Numerator = $z \times (5z^2 - 80) = z \times 5(z - 4)(z + 4) = 5z(z - 4)(z + 4)$.

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The division problem now becomes:

$\frac{5z(z - 4)(z + 4)}{5z(z + 4)}$

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Assume $z \neq 0$ and $z \neq -4$, so that the denominator is not zero.

Cancel the common factors in the numerator and the denominator:

The factor 5 is common.

The factor $z$ is common.

The binomial factor $(z + 4)$ is common.

$\frac{\cancel{5}\cancel{z}(z - 4)\cancel{(z + 4)}}{\cancel{5}\cancel{z}\cancel{(z + 4)}} = z - 4$

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The result of the division is $z - 4$.

(i) $28x^4 \div 56x$

We can write the division as a fraction:

$\frac{28x^4}{56x}$

Divide the numerical coefficients and the variable parts separately:

$\frac{28}{56} = \frac{1}{2}$

$\frac{x^4}{x} = x^{4-1} = x^3$

Multiply the results:

$\frac{1}{2} \times x^3 = \frac{1}{2}x^3$

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(ii) $-36y^3 \div 9y^2$

We can write the division as a fraction:

$\frac{-36y^3}{9y^2}$

Divide the numerical coefficients and the variable parts separately:

$\frac{-36}{9} = -4$

$\frac{y^3}{y^2} = y^{3-2} = y^1 = y$

Multiply the results:

$-4 \times y = -4y$

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(iii) $66pq^2r^3 \div 11qr^2$

We can write the division as a fraction:

$\frac{66pq^2r^3}{11qr^2}$

Divide the numerical coefficients and the variable parts separately:

$\frac{66}{11} = 6$

$\frac{p}{1} = p$

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

$\frac{r^3}{r^2} = r^{3-2} = r^1 = r$

Multiply the results:

$6 \times p \times q \times r = 6pqr$

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(iv) $34x^3y^3z^3 \div 51xy^2z^3$

We can write the division as a fraction:

$\frac{34x^3y^3z^3}{51xy^2z^3}$

Divide the numerical coefficients and the variable parts separately:

$\frac{34}{51} = \frac{2 \times 17}{3 \times 17} = \frac{2}{3}$

$\frac{x^3}{x} = x^{3-1} = x^2$

$\frac{y^3}{y^2} = y^{3-2} = y^1 = y$

$\frac{z^3}{z^3} = z^{3-3} = z^0 = 1$

Multiply the results:

$\frac{2}{3} \times x^2 \times y \times 1 = \frac{2}{3}x^2y$

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(v) $12a^8b^8 \div (– 6a^6b^4)$

We can write the division as a fraction:

$\frac{12a^8b^8}{-6a^6b^4}$

Divide the numerical coefficients and the variable parts separately:

$\frac{12}{-6} = -2$

$\frac{a^8}{a^6} = a^{8-6} = a^2$

$\frac{b^8}{b^4} = b^{8-4} = b^4$

Multiply the results:

$-2 \times a^2 \times b^4 = -2a^2b^4$

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

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(i) $(5x^2 – 6x) \div 3x$

Write as a fraction:

$\frac{5x^2 - 6x}{3x}$

Divide each term in the numerator by the denominator:

$\frac{5x^2}{3x} - \frac{6x}{3x}$

Perform the division for each term:

$\frac{5x^2}{3x} = \frac{5}{3} \times \frac{x^2}{x} = \frac{5}{3}x^{2-1} = \frac{5}{3}x$

$\frac{6x}{3x} = \frac{6}{3} \times \frac{x}{x} = 2 \times x^{1-1} = 2 \times x^0 = 2 \times 1 = 2$

Combine the results:

$\frac{5}{3}x - 2$

So, $(5x^2 – 6x) \div 3x = \frac{5}{3}x - 2$.

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(ii) $(3y^8 – 4y^6 + 5y^4) \div y^4$

Write as a fraction:

$\frac{3y^8 - 4y^6 + 5y^4}{y^4}$

Divide each term in the numerator by the denominator:

$\frac{3y^8}{y^4} - \frac{4y^6}{y^4} + \frac{5y^4}{y^4}$

Perform the division for each term:

$\frac{3y^8}{y^4} = 3 \times y^{8-4} = 3y^4$

$\frac{4y^6}{y^4} = 4 \times y^{6-4} = 4y^2$

$\frac{5y^4}{y^4} = 5 \times y^{4-4} = 5 \times y^0 = 5 \times 1 = 5$

Combine the results:

$3y^4 - 4y^2 + 5$

So, $(3y^8 – 4y^6 + 5y^4) \div y^4 = 3y^4 - 4y^2 + 5$.

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(iii) $8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) \div 4x^2y^2z^2$

Write as a fraction:

$\frac{8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)}{4x^2y^2z^2}$

Divide the numerical factor 8 by 4:

$\frac{\cancel{8}^2(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)}{\cancel{4}_1x^2y^2z^2} = \frac{2(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)}{x^2y^2z^2}$

Divide each term inside the parenthesis by $x^2y^2z^2$ and multiply by 2:

$2 \left( \frac{x^3y^2z^2}{x^2y^2z^2} + \frac{x^2y^3z^2}{x^2y^2z^2} + \frac{x^2y^2z^3}{x^2y^2z^2} \right)$

Perform the division for each term:

$\frac{x^3y^2z^2}{x^2y^2z^2} = x^{3-2}y^{2-2}z^{2-2} = x^1y^0z^0 = x \times 1 \times 1 = x$

$\frac{x^2y^3z^2}{x^2y^2z^2} = x^{2-2}y^{3-2}z^{2-2} = x^0y^1z^0 = 1 \times y \times 1 = y$

$\frac{x^2y^2z^3}{x^2y^2z^2} = x^{2-2}y^{2-2}z^{3-2} = x^0y^0z^1 = 1 \times 1 \times z = z$

Combine the results inside the parenthesis and multiply by 2:

$2(x + y + z)$

So, $8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) \div 4x^2y^2z^2 = 2(x + y + z)$.

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(iv) $(x^3 + 2x^2 +3x) \div 2x$

Write as a fraction:

$\frac{x^3 + 2x^2 + 3x}{2x}$

Divide each term in the numerator by the denominator:

$\frac{x^3}{2x} + \frac{2x^2}{2x} + \frac{3x}{2x}$

Perform the division for each term:

$\frac{x^3}{2x} = \frac{1}{2} \times \frac{x^3}{x} = \frac{1}{2}x^{3-1} = \frac{1}{2}x^2$

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

$\frac{3x}{2x} = \frac{3}{2} \times \frac{x}{x} = \frac{3}{2} \times x^{1-1} = \frac{3}{2} \times x^0 = \frac{3}{2} \times 1 = \frac{3}{2}$

Combine the results:

$\frac{1}{2}x^2 + x + \frac{3}{2}$

So, $(x^3 + 2x^2 +3x) \div 2x = \frac{1}{2}x^2 + x + \frac{3}{2}$.

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(v) $(p^3q^6 – p^6q^3) \div p^3q^3$

Write as a fraction:

$\frac{p^3q^6 - p^6q^3}{p^3q^3}$

Divide each term in the numerator by the denominator:

$\frac{p^3q^6}{p^3q^3} - \frac{p^6q^3}{p^3q^3}$

Perform the division for each term:

$\frac{p^3q^6}{p^3q^3} = p^{3-3}q^{6-3} = p^0q^3 = 1 \times q^3 = q^3$

$\frac{p^6q^3}{p^3q^3} = p^{6-3}q^{3-3} = p^3q^0 = p^3 \times 1 = p^3$

Combine the results:

$q^3 - p^3$

So, $(p^3q^6 – p^6q^3) \div p^3q^3 = q^3 - p^3$.

(i) $(10x – 25) \div 5$

We can write the expression as a fraction:

$\frac{10x - 25}{5}$

Factorise the numerator by taking out the common factor 5:

$10x - 25 = 5(2x - 5)$

Substitute the factorised numerator back into the fraction:

$\frac{5(2x - 5)}{5}$

Cancel the common factor 5 from the numerator and the denominator:

$\frac{\cancel{5}(2x - 5)}{\cancel{5}} = 2x - 5$

The result is $2x - 5$.

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(ii) $(10x – 25) \div (2x – 5)$

We can write the expression as a fraction:

$\frac{10x - 25}{2x - 5}$

Factorise the numerator by taking out the common factor 5:

$10x - 25 = 5(2x - 5)$

Substitute the factorised numerator back into the fraction:

$\frac{5(2x - 5)}{2x - 5}$

Cancel the common binomial factor $(2x - 5)$ from the numerator and the denominator (assuming $2x - 5 \neq 0$):

$\frac{5\cancel{(2x - 5)}}{\cancel{(2x - 5)}} = 5$

The result is 5.

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(iii) $10y(6y + 21) \div 5(2y + 7)$

We can write the expression as a fraction:

$\frac{10y(6y + 21)}{5(2y + 7)}$

Factorise the binomial $(6y + 21)$ in the numerator by taking out the common factor 3:

$6y + 21 = 3(2y + 7)$

Substitute this factorisation back into the numerator:

Numerator $= 10y \times 3(2y + 7) = 30y(2y + 7)$

The fraction becomes:

$\frac{30y(2y + 7)}{5(2y + 7)}$

Cancel the common factors. The numerical factor $\frac{30}{5} = 6$. The binomial factor $(2y + 7)$ is also common (assuming $2y + 7 \neq 0$).

$\frac{\cancel{30}^{\;6}y\cancel{(2y + 7)}}{\cancel{5}_{\;1}\cancel{(2y + 7)}} = 6y$

The result is $6y$.

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(iv) $9x^2y^2(3z – 24) \div 27xy(z – 8)$

We can write the expression as a fraction:

$\frac{9x^2y^2(3z - 24)}{27xy(z - 8)}$

Factorise the binomial $(3z - 24)$ in the numerator by taking out the common factor 3:

$3z - 24 = 3(z - 8)$

Substitute this factorisation back into the numerator:

Numerator $= 9x^2y^2 \times 3(z - 8) = 27x^2y^2(z - 8)$

The fraction becomes:

$\frac{27x^2y^2(z - 8)}{27xy(z - 8)}$

Cancel the common factors. The numerical factor is 27. The variable factors are $x$ (lowest power is 1) and $y$ (lowest power is 1). The binomial factor is $(z - 8)$ (assuming $z - 8 \neq 0$).

$\frac{\cancel{27}\cancel{x^2}^{\;x}\cancel{y^2}^{\;y}\cancel{(z - 8)}}{\cancel{27}\cancel{x}_{\;1}\cancel{y}_{\;1}\cancel{(z - 8)}} = xy$

The result is $xy$.

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(v) $96abc(3a – 12) (5b – 30) \div 144(a – 4) (b – 6)$

We can write the expression as a fraction:

$\frac{96abc(3a - 12)(5b - 30)}{144(a - 4)(b - 6)}$

Factorise the binomial $(3a - 12)$ in the numerator by taking out the common factor 3:

$3a - 12 = 3(a - 4)$

Factorise the binomial $(5b - 30)$ in the numerator by taking out the common factor 5:

$5b - 30 = 5(b - 6)$

Substitute these factorisations back into the numerator:

Numerator $= 96abc \times 3(a - 4) \times 5(b - 6)$

Numerator $= (96 \times 3 \times 5) abc (a - 4)(b - 6) = 1440 abc (a - 4)(b - 6)$

The fraction becomes:

$\frac{1440 abc (a - 4)(b - 6)}{144(a - 4)(b - 6)}$

Cancel the common factors. The numerical factor is $\frac{1440}{144} = 10$. The binomial factors are $(a - 4)$ and $(b - 6)$ (assuming $a - 4 \neq 0$ and $b - 6 \neq 0$).

$\frac{\cancel{1440}^{\;10} abc \cancel{(a - 4)}\cancel{(b - 6)}}{\cancel{144}_{\;1}\cancel{(a - 4)}\cancel{(b - 6)}} = 10abc$

The result is $10abc$.

(i) $5(2x + 1) (3x + 5) \div (2x + 1)$

Write as a fraction:

$\frac{5(2x + 1) (3x + 5)}{(2x + 1)}$

Cancel the common binomial factor $(2x + 1)$ from the numerator and the denominator (assuming $2x + 1 \neq 0$):

$\frac{5\cancel{(2x + 1)} (3x + 5)}{\cancel{(2x + 1)}} = 5(3x + 5)$

The result is $5(3x + 5)$.

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(ii) $26xy(x + 5)(y – 4) \div 13x(y – 4)$

Write as a fraction:

$\frac{26xy(x + 5)(y – 4)}{13x(y – 4)}$

Cancel the common factors. The numerical factor is $\frac{26}{13} = 2$. The variable factor is $x$ (assuming $x \neq 0$). The binomial factor is $(y - 4)$ (assuming $y - 4 \neq 0$).

$\frac{\cancel{26}^{\;2}\cancel{x}y(x + 5)\cancel{(y – 4)}}{\cancel{13}_{\;1}\cancel{x}\cancel{(y – 4)}} = 2y(x + 5)$

The result is $2y(x + 5)$.

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(iii) $52pqr(p + q) (q + r) (r + p) \div 104pq(q + r) (r + p)$

Write as a fraction:

$\frac{52pqr(p + q) (q + r) (r + p)}{104pq(q + r) (r + p)}$

Cancel the common factors. The numerical factor is $\frac{52}{104} = \frac{1}{2}$. The variable factors are $p$ and $q$ (assuming $p \neq 0, q \neq 0$). The binomial factors are $(q + r)$ (assuming $q + r \neq 0$) and $(r + p)$ (assuming $r + p \neq 0$).

$\frac{\cancel{52}^{\;1}\cancel{p}\cancel{q}r(p + q) \cancel{(q + r)} \cancel{(r + p)}}{\cancel{104}_{\;2}\cancel{p}\cancel{q}\cancel{(q + r)} \cancel{(r + p)}} = \frac{1}{2}r(p + q)$

The result is $\frac{1}{2}r(p + q)$.

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(iv) $20(y + 4) (y^2 + 5y + 3) \div 5(y + 4)$

Write as a fraction:

$\frac{20(y + 4) (y^2 + 5y + 3)}{5(y + 4)}$

Cancel the common factors. The numerical factor is $\frac{20}{5} = 4$. The binomial factor is $(y + 4)$ (assuming $y + 4 \neq 0$).

$\frac{\cancel{20}^{\;4}\cancel{(y + 4)} (y^2 + 5y + 3)}{\cancel{5}_{\;1}\cancel{(y + 4)}} = 4(y^2 + 5y + 3)$

The result is $4(y^2 + 5y + 3)$.

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(v) $x(x + 1) (x + 2) (x + 3) \div x(x + 1)$

Write as a fraction:

$\frac{x(x + 1) (x + 2) (x + 3)}{x(x + 1)}$

Cancel the common factors. The variable factor is $x$ (assuming $x \neq 0$). The binomial factor is $(x + 1)$ (assuming $x + 1 \neq 0$).

$\frac{\cancel{x}\cancel{(x + 1)} (x + 2) (x + 3)}{\cancel{x}\cancel{(x + 1)}} = (x + 2)(x + 3)$

The result is $(x + 2)(x + 3)$.

(i) $(y^2 + 7y + 10) \div (y + 5)$

Expression as a fraction: $\frac{y^2 + 7y + 10}{y + 5}$

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Factorise the numerator $y^2 + 7y + 10$. We need two numbers whose product is 10 and sum is 7. These numbers are 2 and 5.

$y^2 + 7y + 10 = (y + 2)(y + 5)$

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Substitute the factorised numerator into the fraction:

$\frac{(y + 2)(y + 5)}{y + 5}$

Cancel the common factor $(y + 5)$ (assuming $y + 5 \neq 0$):

$\frac{(y + 2)\cancel{(y + 5)}}{\cancel{(y + 5)}} = y + 2$

The result is $y + 2$.

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(ii) $(m^2 – 14m – 32) \div (m + 2)$

Expression as a fraction: $\frac{m^2 - 14m - 32}{m + 2}$

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Factorise the numerator $m^2 - 14m - 32$. We need two numbers whose product is -32 and sum is -14. Since the product is negative and the sum is negative, the numbers must have different signs, and the negative number must have a larger absolute value.

Pairs of factors of -32: (1, -32), (-1, 32), (2, -16), (-2, 16), (4, -8), (-4, 8).

Check the sums: $1 + (-32) = -31$, $(-1) + 32 = 31$, $2 + (-16) = -14$. The numbers are 2 and -16.

$m^2 - 14m - 32 = (m + 2)(m - 16)$

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Substitute the factorised numerator into the fraction:

$\frac{(m + 2)(m - 16)}{m + 2}$

Cancel the common factor $(m + 2)$ (assuming $m + 2 \neq 0$):

$\frac{\cancel{(m + 2)}(m - 16)}{\cancel{(m + 2)}} = m - 16$

The result is $m - 16$.

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(iii) $(5p^2 – 25p + 20) \div (p – 1)$

Expression as a fraction: $\frac{5p^2 - 25p + 20}{p - 1}$

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Factorise the numerator $5p^2 – 25p + 20$. First, factor out the common factor 5.

$5p^2 – 25p + 20 = 5(p^2 - 5p + 4)$

Now, factorise the trinomial $p^2 - 5p + 4$. We need two numbers whose product is 4 and sum is -5. Since the product is positive and the sum is negative, both numbers must be negative. These numbers are -1 and -4.

$p^2 - 5p + 4 = (p - 1)(p - 4)$

So, the fully factorised numerator is $5(p - 1)(p - 4)$.

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Substitute the factorised numerator into the fraction:

$\frac{5(p - 1)(p - 4)}{p - 1}$

Cancel the common factor $(p - 1)$ (assuming $p - 1 \neq 0$):

$\frac{5\cancel{(p - 1)}(p - 4)}{\cancel{(p - 1)}} = 5(p - 4)$

The result is $5(p - 4)$.

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(iv) $4yz(z^2 + 6z – 16) \div 2y(z + 8)$

Expression as a fraction: $\frac{4yz(z^2 + 6z - 16)}{2y(z + 8)}$

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Factorise the trinomial $z^2 + 6z – 16$ in the numerator. We need two numbers whose product is -16 and sum is 6. These numbers are 8 and -2.

$z^2 + 6z – 16 = (z + 8)(z - 2)$

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Substitute the factorised trinomial back into the numerator:

Numerator $= 4yz(z + 8)(z - 2)$

The fraction becomes:

$\frac{4yz(z + 8)(z - 2)}{2y(z + 8)}$

Cancel the common factors. The numerical factor is $\frac{4}{2} = 2$. The variable factor is $y$ (assuming $y \neq 0$). The binomial factor is $(z + 8)$ (assuming $z + 8 \neq 0$).

$\frac{\cancel{4}^{\;2}\cancel{y}z\cancel{(z + 8)}(z - 2)}{\cancel{2}_{\;1}\cancel{y}\cancel{(z + 8)}} = 2z(z - 2)$

The result is $2z(z - 2)$.

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(v) $5pq(p^2 – q^2) \div 2p(p + q)$

Expression as a fraction: $\frac{5pq(p^2 - q^2)}{2p(p + q)}$

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Factorise the binomial $p^2 – q^2$ in the numerator using the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$.

$p^2 – q^2 = (p - q)(p + q)$

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Substitute the factorised binomial back into the numerator:

Numerator $= 5pq(p - q)(p + q)$

The fraction becomes:

$\frac{5pq(p - q)(p + q)}{2p(p + q)}$

Cancel the common factors. The numerical factor is $\frac{5}{2}$. The variable factor is $p$ (assuming $p \neq 0$). The binomial factor is $(p + q)$ (assuming $p + q \neq 0$).

$\frac{5\cancel{p}q(p - q)\cancel{(p + q)}}{2\cancel{p}\cancel{(p + q)}} = \frac{5}{2}q(p - q)$

The result is $\frac{5}{2}q(p - q)$.

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(vi) $12xy(9x^2 – 16y^2) \div 4xy(3x + 4y)$

Expression as a fraction: $\frac{12xy(9x^2 - 16y^2)}{4xy(3x + 4y)}$

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Factorise the binomial $9x^2 – 16y^2$ in the numerator using the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$.

$9x^2 = (3x)^2$ and $16y^2 = (4y)^2$.

$9x^2 – 16y^2 = (3x - 4y)(3x + 4y)$

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Substitute the factorised binomial back into the numerator:

Numerator $= 12xy(3x - 4y)(3x + 4y)$

The fraction becomes:

$\frac{12xy(3x - 4y)(3x + 4y)}{4xy(3x + 4y)}$

Cancel the common factors. The numerical factor is $\frac{12}{4} = 3$. The variable factors are $x$ and $y$ (assuming $x \neq 0, y \neq 0$). The binomial factor is $(3x + 4y)$ (assuming $3x + 4y \neq 0$).

$\frac{\cancel{12}^{\;3}\cancel{x}\cancel{y}(3x - 4y)\cancel{(3x + 4y)}}{\cancel{4}_{\;1}\cancel{x}\cancel{y}\cancel{(3x + 4y)}} = 3(3x - 4y)$

The result is $3(3x - 4y)$.

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(vii) $39y^3(50y^2 – 98) \div 26y^2(5y + 7)$

Expression as a fraction: $\frac{39y^3(50y^2 - 98)}{26y^2(5y + 7)}$

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Factorise the binomial $50y^2 – 98$ in the numerator. First, factor out the common numerical factor of 50 and 98. The HCF(50, 98) is 2.

$50y^2 – 98 = 2(25y^2 - 49)$

Now, factorise the expression in the parenthesis, $25y^2 - 49$, which is a difference of squares.

$25y^2 = (5y)^2$ and $49 = 7^2$.

$25y^2 - 49 = (5y - 7)(5y + 7)$

So, the fully factorised form of $50y^2 - 98$ is $2(5y - 7)(5y + 7)$.

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Substitute this factorisation back into the numerator:

Numerator $= 39y^3 \times 2(5y - 7)(5y + 7) = (39 \times 2)y^3(5y - 7)(5y + 7) = 78y^3(5y - 7)(5y + 7)$

The fraction becomes:

$\frac{78y^3(5y - 7)(5y + 7)}{26y^2(5y + 7)}$

Cancel the common factors. The numerical factor is $\frac{78}{26} = 3$. The variable factor is $y^2$ (assuming $y \neq 0$). The binomial factor is $(5y + 7)$ (assuming $5y + 7 \neq 0$).

$\frac{\cancel{78}^{\;3}\cancel{y^3}^{\;y}(5y - 7)\cancel{(5y + 7)}}{\cancel{26}_{\;1}\cancel{y^2}\cancel{(5y + 7)}} = 3y(5y - 7)$

The result is $3y(5y - 7)$.