Chapter 14 Factorisation

We are asked to factorise the expression $12a^2b + 15ab^2$.

To factorise an expression, we find the greatest common factor (GCF) of the terms and factor it out.

The terms are $12a^2b$ and $15ab^2$.

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Find the GCF of the numerical coefficients 12 and 15.

Prime factorization of 12: $2 \times 2 \times 3 = 2^2 \times 3^1$

Prime factorization of 15: $3 \times 5$

The common prime factor is 3. The lowest power of 3 is $3^1$.

GCF of 12 and 15 is 3.

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Find the GCF of the variable parts $a^2b$ and $ab^2$.

For the variable $a$: The terms have $a^2$ and $a^1$. The lowest power is $a^1$, which is $a$.

For the variable $b$: The terms have $b^1$ and $b^2$. The lowest power is $b^1$, which is $b$.

GCF of the variable parts is $ab$.

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The GCF of the entire expression $12a^2b + 15ab^2$ is the product of the GCF of numerical and variable parts.

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

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Now, factor out the GCF from each term:

Divide the first term by the GCF: $\frac{12a^2b}{3ab} = \frac{12}{3} \times \frac{a^2}{a} \times \frac{b}{b} \ $$ = 4 \times a^{2-1} \times b^{1-1} = 4a^1 b^0 = 4a$.

Divide the second term by the GCF: $\frac{15ab^2}{3ab} = \frac{15}{3} \times \frac{a}{a} \times \frac{b^2}{b} \ $$ = 5 \times a^{1-1} \times b^{2-1} = 5a^0 b^1 = 5b$.

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Write the factored expression as the GCF multiplied by the sum of the results from the division:

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

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The factored form of the expression is $3ab(4a + 5b)$.

We are asked to factorise the expression $10x^2 - 18x^3 + 14x^4$.

To factorise a polynomial, we find the greatest common factor (GCF) of all its terms and factor it out.

The terms are $10x^2$, $-18x^3$, and $14x^4$.

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Find the GCF of the numerical coefficients 10, -18, and 14. We consider the positive GCF of 10, 18, and 14.

Prime factorization of 10 is $2 \times 5$.

Prime factorization of 18 is $2 \times 3^2$.

Prime factorization of 14 is $2 \times 7$.

The common prime factor is 2, and its lowest power is $2^1$.

The GCF of 10, 18, and 14 is 2.

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Find the GCF of the variable parts $x^2$, $x^3$, and $x^4$.

The variable is $x$, and the powers are 2, 3, and 4. The lowest power is 2.

The GCF of the variable parts is $x^2$.

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

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

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Now, factor out the GCF $2x^2$ from each term by dividing each term by the GCF:

$\frac{10x^2}{2x^2} = 5$

$\frac{-18x^3}{2x^2} = -9x$

$\frac{14x^4}{2x^2} = 7x^2$

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Write the factored expression as the GCF multiplied by the sum/difference of the results from the division:

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

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The factored form of the expression is $2x^2(5 - 9x + 7x^2)$.

We are asked to factorise the expression $6xy – 4y + 6 – 9x$.

This is a polynomial with four terms. We can try factoring by grouping.

Group the first two terms and the last two terms:

$(6xy – 4y) + (6 – 9x)$

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Factor out the greatest common factor (GCF) from the first group $(6xy – 4y)$.

The GCF of $6xy$ and $4y$ is $2y$.

$6xy – 4y = 2y(\frac{6xy}{2y} - \frac{4y}{2y}) = 2y(3x - 2)$.

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Factor out the GCF from the second group $(6 – 9x)$.

The GCF of 6 and $9x$ is 3.

$6 – 9x = 3(\frac{6}{3} - \frac{9x}{3}) = 3(2 - 3x)$.

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The expression now is $2y(3x - 2) + 3(2 - 3x)$.

Notice that the binomial factor $(2 - 3x)$ is the opposite of $(3x - 2)$.

We can write $(2 - 3x)$ as $-(3x - 2)$.

Substitute this into the expression:

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

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

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

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The factored form of the expression is $(3x - 2)(2y - 3)$.

To find the common factors of given terms, we find the greatest common factor (GCF) of the numerical coefficients and the lowest power of each common variable.

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(i) Terms: $12x$ and $36$.

GCF of numerical coefficients (12, 36): $\text{GCF}(12, 36) = 12$.

Variables: $x$ and no variable. The only common factor for variables is 1.

Common factor = $12 \times 1 = 12$.

The common factor is 12.

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(ii) Terms: $2y$ and $22xy$.

GCF of numerical coefficients (2, 22): $\text{GCF}(2, 22) = 2$.

Variables: $y$ and $xy$. The common variable is $y$. The lowest power of $y$ is $y^1$, which is $y$.

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

The common factor is $2y$.

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(iii) Terms: $14pq$ and $28p^2q^2$.

GCF of numerical coefficients (14, 28): $\text{GCF}(14, 28) = 14$.

Variables: $pq$ ($p^1q^1$) and $p^2q^2$. The common variables are $p$ and $q$.

Lowest power of $p$: $p^1$.

Lowest power of $q$: $q^1$.

GCF of variable parts = $p^1q^1 = pq$.

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

The common factor is $14pq$.

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(iv) Terms: $2x$, $3x^2$, and $4$.

GCF of numerical coefficients (2, 3, 4): $\text{GCF}(2, 3, 4) = 1$.

Variables: $x$, $x^2$, and no variable. There is no variable common to all three terms.

Common factor = $1 \times 1 = 1$.

The common factor is 1.

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(v) Terms: $6abc$, $24ab^2$, and $12a^2b$.

GCF of numerical coefficients (6, 24, 12): $\text{GCF}(6, 24, 12) = 6$.

Variables: $abc$ ($a^1b^1c^1$), $ab^2$ ($a^1b^2$), and $a^2b$ ($a^2b^1$). The common variables are $a$ and $b$. The variable $c$ is not common to all.

Lowest power of $a$: $a^1$.

Lowest power of $b$: $b^1$.

GCF of variable parts = $a^1b^1 = ab$.

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

The common factor is $6ab$.

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(vi) Terms: $16x^3$, $-4x^2$, and $32x$.

GCF of numerical coefficients (16, 4, 32) considering positive value: $\text{GCF}(16, 4, 32) = 4$.

Variables: $x^3$, $x^2$, and $x$ ($x^1$). The common variable is $x$.

Lowest power of $x$: $x^1$.

GCF of variable parts = $x^1 = x$.

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

The common factor is $4x$.

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(vii) Terms: $10pq$, $20qr$, and $30rp$.

GCF of numerical coefficients (10, 20, 30): $\text{GCF}(10, 20, 30) = 10$.

Variables: $pq$, $qr$, and $rp$. No single variable ($p$, $q$, or $r$) appears in all three terms.

GCF of variable parts = 1.

Common factor = $10 \times 1 = 10$.

The common factor is 10.

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(viii) Terms: $3x^2y^3$, $10x^3y^2$, and $6x^2y^2z$.

GCF of numerical coefficients (3, 10, 6): $\text{GCF}(3, 10, 6) = 1$.

Variables: $x^2y^3$, $x^3y^2$, and $x^2y^2z$. The common variables are $x$ and $y$. The variable $z$ is not common to all.

For variable $x$: powers are 2, 3, 2. Lowest power is $x^2$.

For variable $y$: powers are 3, 2, 2. Lowest power is $y^2$.

GCF of variable parts = $x^2y^2$.

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

The common factor is $x^2y^2$.

To factorise an expression, we find the greatest common factor (GCF) of the terms and factor it out.

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(i) Expression: $7x – 42$.

The terms are $7x$ and $-42$.

The GCF of the numerical coefficients 7 and 42 is 7.

There is no common variable.

GCF $= 7$.

Factor out the GCF: $7x - 42 = 7(\frac{7x}{7} - \frac{42}{7}) = 7(x - 6)$.

The factored form is $7(x - 6)$.

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(ii) Expression: $6p – 12q$.

The terms are $6p$ and $-12q$.

The GCF of the numerical coefficients 6 and 12 is 6.

There is no common variable ($p$ and $q$ are different).

GCF $= 6$.

Factor out the GCF: $6p - 12q = 6(\frac{6p}{6} - \frac{12q}{6}) = 6(p - 2q)$.

The factored form is $6(p - 2q)$.

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(iii) Expression: $7a^2 + 14a$.

The terms are $7a^2$ and $14a$.

The GCF of the numerical coefficients 7 and 14 is 7.

The common variable is $a$. The lowest power of $a$ is $a^1$.

GCF $= 7a$.

Factor out the GCF: $7a^2 + 14a = 7a(\frac{7a^2}{7a} + \frac{14a}{7a}) = 7a(a + 2)$.

The factored form is $7a(a + 2)$.

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(iv) Expression: –16z + 20z³.

The terms are $-16z$ and $20z^3$.

The GCF of the absolute values of numerical coefficients (16, 20) is 4.

The common variable is $z$. The lowest power of $z$ is $z^1$.

GCF $= 4z$. (We can also factor out $-4z$, but $4z$ is also correct).

Factor out the GCF: $-16z + 20z^3 = 4z(\frac{-16z}{4z} + \frac{20z^3}{4z}) = 4z(-4 + 5z^2)$.

The factored form is $4z(-4 + 5z^2)$ or $4z(5z^2 - 4)$.

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

The terms are $20l^2m$ and $30alm$.

The GCF of the numerical coefficients 20 and 30 is 10.

Common variables are $l$ and $m$. Lowest power of $l$ is $l^1$, lowest power of $m$ is $m^1$. The variable $a$ is not common.

GCF $= 10lm$.

Factor out the GCF: $20l^2m + 30alm = 10lm(\frac{20l^2m}{10lm} + \frac{30alm}{10lm}) \ $$ = 10lm(2l + 3a)$.

The factored form is $10lm(2l + 3a)$.

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(vi) Expression: $5x^2y – 15xy^2$.

The terms are $5x^2y$ and $-15xy^2$.

The GCF of the numerical coefficients 5 and 15 is 5.

Common variables are $x$ and $y$. Lowest power of $x$ is $x^1$, lowest power of $y$ is $y^1$.

GCF $= 5xy$.

Factor out the GCF: $5x^2y - 15xy^2 = 5xy(\frac{5x^2y}{5xy} - \frac{15xy^2}{5xy}) = 5xy(x - 3y)$.

The factored form is $5xy(x - 3y)$.

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(vii) Expression: $10a^2 – 15b^2 + 20c^2$.

The terms are $10a^2$, $-15b^2$, and $20c^2$.

The GCF of the numerical coefficients 10, 15, and 20 is 5.

There are no variables common to all three terms ($a$, $b$, $c$ are distinct).

GCF $= 5$.

Factor out the GCF: $10a^2 - 15b^2 + 20c^2 = 5(\frac{10a^2}{5} - \frac{15b^2}{5} + \frac{20c^2}{5}) \ $$ = 5(2a^2 - 3b^2 + 4c^2)$.

The factored form is $5(2a^2 - 3b^2 + 4c^2)$.

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(viii) Expression: – 4a² + 4ab – 4ca.

The terms are $-4a^2$, $4ab$, and $-4ca$.

The GCF of the absolute values of numerical coefficients (4, 4, 4) is 4.

The common variable is $a$. The lowest power of $a$ is $a^1$.

GCF $= 4a$. (We can also factor out $-4a$).

Factor out the GCF $4a$: $-4a^2 + 4ab - 4ca = 4a(\frac{-4a^2}{4a} + \frac{4ab}{4a} - \frac{4ca}{4a}) \ $$ = 4a(-a + b - c)$.

The factored form is $4a(-a + b - c)$ or $4a(b - a - c)$.

If we factor out $-4a$: $-4a^2 + 4ab - 4ca = -4a(\frac{-4a^2}{-4a} + \frac{4ab}{-4a} - \frac{4ca}{-4a}) \ $$ = -4a(a - b + c)$.

The factored form is also $-4a(a - b + c)$. Both forms are correct.

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

The terms are $x^2yz$, $xy^2z$, and $xyz^2$.

There are no numerical coefficients other than 1.

Common variables are $x$, $y$, and $z$.

Lowest power of $x$: $x^1$.

Lowest power of $y$: $y^1$.

Lowest power of $z$: $z^1$.

GCF $= xyz$.

Factor out the GCF: $x^2yz + xy^2z + xyz^2 = xyz(\frac{x^2yz}{xyz} + \frac{xy^2z}{xyz} + \frac{xyz^2}{xyz}) \ $$ = xyz(x + y + z)$.

The factored form is $xyz(x + y + z)$.

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(x) Expression: $ax^2y + bxy^2 + cxyz$.

The terms are $ax^2y$, $bxy^2$, and $cxyz$.

Assuming $a, b, c$ are constants, the GCF of numerical parts is 1.

Common variables are $x$ and $y$. The variable $z$ is not common to all terms.

Lowest power of $x$: $x^1$.

Lowest power of $y$: $y^1$.

GCF $= xy$.

Factor out the GCF: $ax^2y + bxy^2 + cxyz = xy(\frac{ax^2y}{xy} + \frac{bxy^2}{xy} + \frac{cxyz}{xy}) \ $$ = xy(ax + by + cz)$.

The factored form is $xy(ax + by + cz)$.

To factorise these expressions, we will use the method of grouping terms.

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

Group the first two terms and the last two terms:

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

Factor out the greatest common factor (GCF) from each group:

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

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

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

The factored form is $(x + y)(x + 8)$.

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(ii) Factorise $15xy – 6x + 5y – 2$.

Group the first two terms and the last two terms:

$(15xy – 6x) + (5y – 2)$

Factor out the GCF from each group:

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

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

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

The factored form is $(5y - 2)(3x + 1)$.

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(iii) Factorise $ax + bx – ay – by$.

Group the first two terms and the last two terms:

$(ax + bx) + (-ay – by)$

Factor out the GCF from each group:

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

Notice that $(a + b)$ is a common binomial factor. Factor it out:

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

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

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

Rearrange and group terms to find common factors. Let's group terms with $q$ and terms with numerical factors.

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

Factor out the GCF from each group:

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

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

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

The factored form is $(5p + 3)(3q + 5)$.

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(v) Factorise $z – 7 + 7xy – xyz$.

Group the first two terms and the last two terms:

$(z – 7) + (7xy – xyz)$

Factor out the GCF from each group:

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

Notice that the binomial factors $(z - 7)$ and $(7 - z)$ are opposites. We can write $(7 - z)$ as $-(z - 7)$.

The expression becomes:

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

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

Notice that $(z - 7)$ is now a common binomial factor. Factor it out:

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

The factored form is $(z - 7)(1 - xy)$.

We are asked to factorise the expression $x^2 + 8x + 16$.

This is a quadratic trinomial. We can try to express it using an algebraic identity.

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Consider the identity $(a+b)^2 = a^2 + 2ab + b^2$.

Compare the given expression $x^2 + 8x + 16$ with the right side of the identity $a^2 + 2ab + b^2$.

Let $a^2 = x^2$, which implies $a = x$.

Let $b^2 = 16$, which implies $b = 4$ (assuming $b$ is positive).

Now, check the middle term $2ab$:

$2ab = 2 \times x \times 4 = 8x$.

This matches the middle term of the given expression.

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Since $x^2 + 8x + 16$ is in the form $a^2 + 2ab + b^2$ with $a=x$ and $b=4$, we can write it as $(a+b)^2$.

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

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The factored form of the expression is $(x + 4)^2$.

We are asked to factorise the expression $4y^2 – 12y + 9$.

This is a quadratic trinomial. We can try to express it using an algebraic identity.

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Consider the identity $(a-b)^2 = a^2 - 2ab + b^2$.

Compare the given expression $4y^2 – 12y + 9$ with the right side of the identity $a^2 - 2ab + b^2$.

Let $a^2 = 4y^2$, which implies $a = \sqrt{4y^2} = 2y$ (assuming $a$ is positive).

Let $b^2 = 9$, which implies $b = \sqrt{9} = 3$ (assuming $b$ is positive).

Now, check the middle term $-2ab$:

$-2ab = -2 \times (2y) \times 3 = -12y$.

This matches the middle term of the given expression.

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Since $4y^2 – 12y + 9$ is in the form $a^2 - 2ab + b^2$ with $a=2y$ and $b=3$, we can write it as $(a-b)^2$.

$4y^2 – 12y + 9 = (2y - 3)^2$.

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The factored form of the expression is $(2y - 3)^2$.

We are asked to factorise the expression $49p^2 – 36$.

This expression is in the form of a difference of two perfect squares.

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Recall the algebraic identity for the difference of squares:

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

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Compare the given expression $49p^2 - 36$ with $a^2 - b^2$.

The first term is $49p^2$. We can write $49p^2$ as $(7p)^2$.

So, $a^2 = 49p^2 \implies a = 7p$.

The second term is $36$. We can write $36$ as $6^2$.

So, $b^2 = 36 \implies b = 6$.

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Now, apply the identity $a^2 - b^2 = (a + b)(a - b)$ with $a = 7p$ and $b = 6$:

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

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The factored form of the expression is $(7p + 6)(7p - 6)$.

We are asked to factorise the expression $a^2 – 2ab + b^2 – c^2$.

Let's look at the first three terms: $a^2 – 2ab + b^2$. This is a perfect square trinomial, which is the expansion of $(a - b)^2$.

Recall the identity: $(a-b)^2 = a^2 - 2ab + b^2$.

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Substitute this into the given expression:

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

$= (a - b)^2 – c^2$.

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Now, the expression is in the form of a difference of two perfect squares, $X^2 - Y^2$, where $X = (a-b)$ and $Y = c$.

Recall the identity: $X^2 - Y^2 = (X + Y)(X - Y)$.

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Apply this identity by substituting $X = (a-b)$ and $Y = c$:

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

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

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

We are asked to factorise the expression $m^4 – 256$.

We can recognise this expression as a difference of two terms.

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We can rewrite each term as a perfect square:

$m^4 = (m^2)^2$

$256 = 16^2$

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

Recall the algebraic identity: $a^2 - b^2 = (a + b)(a - b)$.

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Apply the identity with $a = m^2$ and $b = 16$:

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

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Now, consider the factors $(m^2 + 16)$ and $(m^2 - 16)$.

The factor $(m^2 + 16)$ is a sum of squares and cannot be factorised further using real numbers.

The factor $(m^2 - 16)$ is another difference of two perfect squares, $x^2 - y^2$, where $x = m$ and $y = 4$ (since $m^2 = m^2$ and $16 = 4^2$).

Apply the identity $x^2 - y^2 = (x + y)(x - y)$ to $(m^2 - 16)$:

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

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Substitute this back into the factorised expression:

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

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The completely factored form of the expression is $(m^2 + 16)(m + 4)(m - 4)$.

We are asked to factorise the quadratic trinomial $x^2 + 5x + 6$.

This expression is in the form $x^2 + bx + c$, where $b=5$ and $c=6$.

To factorise this type of trinomial, we look for two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the $x$ term ($p+q = 5$) and their product is equal to the constant term ($p \times q = 6$).

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

$p + q = 5$

$p \times q = 6$

Let's list pairs of integers whose product is 6:

(1, 6), (-1, -6), (2, 3), (-2, -3).

Now, let's check the sum of each pair:

$1 + 6 = 7$

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

$2 + 3 = 5$

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

The pair of numbers that satisfy both conditions are 2 and 3.

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Now, we can rewrite the middle term $5x$ as the sum of $2x$ and $3x$:

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

Now, we can factor by grouping the terms:

Group the first two terms and the last two terms:

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

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Factor out the greatest common factor (GCF) from the first group $(x^2 + 2x)$:

$x(x + 2)$.

Factor out the GCF from the second group $(3x + 6)$:

$3(x + 2)$.

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The expression is now $x(x + 2) + 3(x + 2)$.

Notice that $(x + 2)$ is a common binomial factor. Factor it out:

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

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The factored form of the expression is $(x + 2)(x + 3)$.

We are asked to find the factors of the quadratic trinomial $y^2 - 7y + 12$.

This expression is in the form $y^2 + by + c$, where $b=-7$ and $c=12$.

To factorise this type of trinomial, we look for two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the $y$ term ($p+q = -7$) and their product is equal to the constant term ($p \times q = 12$).

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

$p + q = -7$

$p \times q = 12$

Let's list pairs of integers whose product is 12:

(1, 12), (-1, -12), (2, 6), (-2, -6), (3, 4), (-3, -4).

Now, let's check the sum of each pair:

$1 + 12 = 13$

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

$2 + 6 = 8$

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

$3 + 4 = 7$

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

The pair of numbers that satisfy both conditions are -3 and -4.

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Now, we can rewrite the middle term $-7y$ as the sum of $-3y$ and $-4y$:

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

Now, we can factor by grouping the terms:

Group the first two terms and the last two terms:

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

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Factor out the greatest common factor (GCF) from the first group $(y^2 - 3y)$:

$y(y - 3)$.

Factor out the GCF from the second group $(-4y + 12)$. The GCF of $-4y$ and 12 is $-4$.

$-4(y - 3)$.

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The expression is now $y(y - 3) - 4(y - 3)$.

Notice that $(y - 3)$ is a common binomial factor. Factor it out:

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

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The factored form of the expression is $(y - 3)(y - 4)$.

We are asked to obtain the factors of the quadratic trinomial $z^2 – 4z – 12$.

This expression is in the form $z^2 + bz + c$, where $b=-4$ and $c=-12$.

To factorise this type of trinomial, we look for two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the $z$ term ($p+q = -4$) and their product is equal to the constant term ($p \times q = -12$).

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

$p + q = -4$

$p \times q = -12$

Let's list pairs of integers whose product is -12:

(1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4).

Now, let's check the sum of each pair:

$1 + (-12) = -11$

$-1 + 12 = 11$

$2 + (-6) = -4$

$-2 + 6 = 4$

$3 + (-4) = -1$

$-3 + 4 = 1$

The pair of numbers that satisfy both conditions are 2 and -6.

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Now, we can rewrite the middle term $-4z$ as the sum of $2z$ and $-6z$:

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

Now, we can factor by grouping the terms:

Group the first two terms and the last two terms:

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

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Factor out the greatest common factor (GCF) from the first group $(z^2 + 2z)$:

$z(z + 2)$.

Factor out the GCF from the second group $(-6z - 12)$. The GCF of $-6z$ and $-12$ is $-6$.

$-6(z + 2)$.

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The expression is now $z(z + 2) - 6(z + 2)$.

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

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

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The factored form of the expression is $(z + 2)(z - 6)$.

We are asked to find the factors of the expression $3m^2 + 9m + 6$.

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First, check if there is a common factor for all the terms in the expression.

The numerical coefficients are 3, 9, and 6.

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

There are no variables common to all terms.

So, the GCF of the expression is 3. Factor out the GCF:

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

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

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

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

We look for two numbers, $p$ and $q$, such that their sum is the coefficient of the middle term ($p+q = 3$) and their product is the constant term ($p \times q = 2$).

The pairs of integers whose product is 2 are (1, 2) and (-1, -2).

Check their sums:

$1 + 2 = 3$. This pair works.

$-1 + (-2) = -3$.

So, the two numbers are 1 and 2.

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Rewrite the middle term $3m$ as $1m + 2m$:

$m^2 + 3m + 2 = m^2 + m + 2m + 2$.

Factor by grouping:

$(m^2 + m) + (2m + 2)$

Factor out the GCF from each group:

$m(m + 1) + 2(m + 1)$.

Factor out the common binomial factor $(m + 1)$:

$(m + 1)(m + 2)$.

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Combine the initial GCF with the factored trinomial:

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

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The factors of the expression $3m^2 + 9m + 6$ are $3$, $(m + 1)$, and $(m + 2)$. The factored form is $3(m + 1)(m + 2)$.

We will use the identities for perfect square trinomials:

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

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

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

This is in the form $x^2 + 2xy + y^2$.

Here, $x^2 = a^2 \implies x = a$.

And $y^2 = 16 \implies y = 4$.

Check the middle term: $2xy = 2(a)(4) = 8a$. This matches the given middle term.

So, $a^2 + 8a + 16 = (a + 4)^2$.

The factored form is $(a + 4)^2$.

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(ii) Factorise $p^2 – 10p + 25$.

This is in the form $x^2 - 2xy + y^2$.

Here, $x^2 = p^2 \implies x = p$.

And $y^2 = 25 \implies y = 5$.

Check the middle term: $-2xy = -2(p)(5) = -10p$. This matches the given middle term.

So, $p^2 – 10p + 25 = (p - 5)^2$.

The factored form is $(p - 5)^2$.

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

This is in the form $x^2 + 2xy + y^2$.

Here, $x^2 = 25m^2 = (5m)^2 \implies x = 5m$.

And $y^2 = 9 = 3^2 \implies y = 3$.

Check the middle term: $2xy = 2(5m)(3) = 30m$. This matches the given middle term.

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

The factored form is $(5m + 3)^2$.

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

This is in the form $x^2 + 2xy + y^2$.

Here, $x^2 = 49y^2 = (7y)^2 \implies x = 7y$.

And $y^2 = 36z^2 = (6z)^2 \implies y = 6z$.

Check the middle term: $2xy = 2(7y)(6z) = 84yz$. This matches the given middle term.

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

The factored form is $(7y + 6z)^2$.

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(v) Factorise $4x^2 – 8x + 4$.

First, notice the common factor 4 in all terms. Factor it out:

$4x^2 – 8x + 4 = 4(x^2 – 2x + 1)$.

Now, consider the trinomial inside the parentheses: $x^2 – 2x + 1$.

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

Here, $a^2 = x^2 \implies a = x$.

And $b^2 = 1 = 1^2 \implies b = 1$.

Check the middle term: $-2ab = -2(x)(1) = -2x$. This matches the middle term.

So, $x^2 – 2x + 1 = (x - 1)^2$.

Combine the factored out 4 with the factored trinomial:

$4(x^2 – 2x + 1) = 4(x - 1)^2$.

The factored form is $4(x - 1)^2$.

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(vi) Factorise $121b^2 – 88bc + 16c^2$.

This is in the form $x^2 - 2xy + y^2$.

Here, $x^2 = 121b^2 = (11b)^2 \implies x = 11b$.

And $y^2 = 16c^2 = (4c)^2 \implies y = 4c$.

Check the middle term: $-2xy = -2(11b)(4c) = -88bc$. This matches the given middle term.

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

The factored form is $(11b - 4c)^2$.

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(vii) Factorise $(l + m)^2 – 4lm$.

First, expand $(l + m)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

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

Now, substitute this back into the expression:

$(l^2 + 2lm + m^2) – 4lm$.

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

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

This resulting expression $l^2 - 2lm + m^2$ is a perfect square trinomial in the form $a^2 - 2ab + b^2$, where $a=l$ and $b=m$.

So, $l^2 - 2lm + m^2 = (l - m)^2$.

The factored form is $(l - m)^2$.

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

We can rewrite the terms to match the identity for a perfect square trinomial.

$a^4 = (a^2)^2$.

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

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

This is in the form $x^2 + 2xy + y^2$, where $x = a^2$ and $y = b^2$.

Check the middle term: $2xy = 2(a^2)(b^2) = 2a^2b^2$. This matches the given middle term.

So, $a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2$.

The factored form is $(a^2 + b^2)^2$.

We will use the identity for the difference of squares: $a^2 - b^2 = (a + b)(a - b)$, and also identify common factors.

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(i) Factorise $4p^2 – 9q^2$.

This is in the form of a difference of squares.

$4p^2 = (2p)^2$

$9q^2 = (3q)^2$

So, $4p^2 – 9q^2 = (2p)^2 – (3q)^2$.

Using $a^2 - b^2 = (a+b)(a-b)$ with $a=2p$ and $b=3q$:

$(2p)^2 – (3q)^2 = (2p + 3q)(2p - 3q)$.

The factored form is $(2p + 3q)(2p - 3q)$.

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(ii) Factorise $63a^2 – 112b^2$.

First, find the greatest common factor (GCF) of 63 and 112.

Prime factorization of 63: $3^2 \times 7$.

Prime factorization of 112: $2^4 \times 7$.

The GCF is 7.

Factor out the GCF:

$63a^2 – 112b^2 = 7(9a^2 – 16b^2)$.

Now, consider the expression in the parentheses: $9a^2 – 16b^2$. This is a difference of squares.

$9a^2 = (3a)^2$

$16b^2 = (4b)^2$

So, $9a^2 – 16b^2 = (3a)^2 – (4b)^2$.

Using $a^2 - b^2 = (a+b)(a-b)$ with $a=3a$ and $b=4b$:

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

Combine the GCF with the factored expression:

$7(9a^2 – 16b^2) = 7(3a + 4b)(3a - 4b)$.

The factored form is $7(3a + 4b)(3a - 4b)$.

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(iii) Factorise $49x^2 – 36$.

This is in the form of a difference of squares.

$49x^2 = (7x)^2$

$36 = 6^2$

So, $49x^2 – 36 = (7x)^2 – 6^2$.

Using $a^2 - b^2 = (a+b)(a-b)$ with $a=7x$ and $b=6$:

$(7x)^2 – 6^2 = (7x + 6)(7x - 6)$.

The factored form is $(7x + 6)(7x - 6)$.

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(iv) Factorise $16x^5 – 144x^3$.

First, find the greatest common factor (GCF) of the terms $16x^5$ and $144x^3$.

GCF of numerical coefficients (16, 144): $\text{GCF}(16, 144)$. Since $144 = 9 \times 16$, the GCF is 16.

GCF of variable parts ($x^5$, $x^3$): The lowest power is $x^3$.

GCF $= 16x^3$.

Factor out the GCF:

$16x^5 – 144x^3 = 16x^3(\frac{16x^5}{16x^3} – \frac{144x^3}{16x^3})$

$16x^5 – 144x^3 = 16x^3(x^2 – 9)$.

Now, consider the expression in the parentheses: $x^2 – 9$. This is a difference of squares.

$x^2 = x^2$

$9 = 3^2$

So, $x^2 – 9 = x^2 – 3^2$.

Using $a^2 - b^2 = (a+b)(a-b)$ with $a=x$ and $b=3$:

$x^2 – 3^2 = (x + 3)(x - 3)$.

Combine the GCF with the factored expression:

$16x^3(x^2 – 9) = 16x^3(x + 3)(x - 3)$.

The factored form is $16x^3(x + 3)(x - 3)$.

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(v) Factorise $(l + m)^2 – (l – m)^2$.

This is directly in the form of a difference of squares, $A^2 - B^2$, where $A = (l+m)$ and $B = (l-m)$.

Using the identity $A^2 - B^2 = (A+B)(A-B)$:

Substitute $A = (l+m)$ and $B = (l-m)$:

$((l + m) + (l – m))((l + m) – (l – m))$

Simplify each part:

First part: $(l + m) + (l – m) = l + m + l - m = 2l$.

Second part: $(l + m) – (l – m) = l + m - l + m = 2m$.

Multiply the simplified parts:

$(2l)(2m) = 4lm$.

The factored form is $4lm$.

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(vi) Factorise $9x^2y^2 – 16$.

This is in the form of a difference of squares.

$9x^2y^2 = (3xy)^2$

$16 = 4^2$

So, $9x^2y^2 – 16 = (3xy)^2 – 4^2$.

Using $a^2 - b^2 = (a+b)(a-b)$ with $a=3xy$ and $b=4$:

$(3xy)^2 – 4^2 = (3xy + 4)(3xy - 4)$.

The factored form is $(3xy + 4)(3xy - 4)$.

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(vii) Factorise $(x^2 – 2xy + y^2) – z^2$.

The expression inside the first parenthesis is a perfect square trinomial: $x^2 – 2xy + y^2 = (x - y)^2$.

Substitute this into the expression:

$(x - y)^2 – z^2$.

This is now in the form of a difference of squares, $A^2 - B^2$, where $A = (x-y)$ and $B = z$.

Using the identity $A^2 - B^2 = (A+B)(A-B)$:

Substitute $A = (x-y)$ and $B = z$:

$((x - y) + z)((x - y) - z)$.

The factored form is $(x - y + z)(x - y - z)$.

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(viii) Factorise $25a^2 – 4b^2 + 28bc – 49c^2$.

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

$25a^2 – (4b^2 – 28bc + 49c^2)$.

Notice the expression inside the parenthesis: $4b^2 – 28bc + 49c^2$. This looks like a perfect square trinomial $(X - Y)^2 = X^2 - 2XY + Y^2$.

Here, $X^2 = 4b^2 = (2b)^2 \implies X = 2b$.

And $Y^2 = 49c^2 = (7c)^2 \implies Y = 7c$.

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

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

Substitute this back into the main expression:

$25a^2 – (2b - 7c)^2$.

This is now in the form of a difference of squares, $A^2 - B^2$, where $A = 5a$ (since $25a^2 = (5a)^2$) and $B = (2b - 7c)$.

Using the identity $A^2 - B^2 = (A+B)(A-B)$:

Substitute $A = 5a$ and $B = (2b - 7c)$:

$(5a + (2b - 7c))(5a - (2b - 7c))$.

Simplify the terms inside the second parenthesis (distribute the minus sign):

$(5a + 2b - 7c)(5a - 2b + 7c)$.

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

To factorise the expressions, we will primarily use the method of finding the greatest common factor (GCF) and grouping terms.

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(i) Factorise $ax^2 + bx$.

Find the GCF of the terms $ax^2$ and $bx$.

The common variable is $x$. The lowest power of $x$ is $x^1$.

There is no common numerical factor or other common variables ($a$ and $b$ are generally treated as distinct constants here).

GCF $= x$.

Factor out the GCF: $ax^2 + bx = x(\frac{ax^2}{x} + \frac{bx}{x}) = x(ax + b)$.

The factored form is $x(ax + b)$.

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

Find the GCF of the terms $7p^2$ and $21q^2$.

The GCF of the numerical coefficients 7 and 21 is 7.

There are no common variables ($p$ and $q$ are distinct).

GCF $= 7$.

Factor out the GCF: $7p^2 + 21q^2 = 7(\frac{7p^2}{7} + \frac{21q^2}{7}) = 7(p^2 + 3q^2)$.

The factored form is $7(p^2 + 3q^2)$.

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

Find the GCF of the terms $2x^3$, $2xy^2$, and $2xz^2$.

The GCF of the numerical coefficients 2, 2, and 2 is 2.

The common variable is $x$. The lowest power of $x$ is $x^1$.

The variables $y$ and $z$ are not common to all terms.

GCF $= 2x$.

Factor out the GCF: $2x^3 + 2xy^2 + 2xz^2 = 2x(\frac{2x^3}{2x} + \frac{2xy^2}{2x} + \frac{2xz^2}{2x}) \ $$ = 2x(x^2 + y^2 + z^2)$.

The factored form is $2x(x^2 + y^2 + z^2)$.

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

This is a four-term polynomial. We can try grouping terms.

Group the first two terms and the last two terms:

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

Factor out the GCF from each group:

From $(am^2 + bm^2)$, GCF is $m^2$: $m^2(a + b)$.

From $(bn^2 + an^2)$, GCF is $n^2$: $n^2(b + a)$, which is the same as $n^2(a + b)$.

The expression becomes $m^2(a + b) + n^2(a + b)$.

Notice the common binomial factor $(a + b)$. Factor it out:

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

The factored form is $(a + b)(m^2 + n^2)$.

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

This is already grouped as two sets of terms.

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

Factor out the GCF from the first group $(lm + l)$:

$l(m + 1)$.

The second group $(m + 1)$ can be written as $1(m + 1)$.

The expression becomes $l(m + 1) + 1(m + 1)$.

Notice the common binomial factor $(m + 1)$. Factor it out:

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

The factored form is $(m + 1)(l + 1)$.

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

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

Factor out the common factor $(y + z)$:

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

The factored form is $(y + z)(y + 9)$.

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(vii) Factorise $5y^2 – 20y – 8z + 2yz$.

This is a four-term polynomial. Let's try grouping.

Group the first two terms and the last two terms:

$(5y^2 – 20y) + (-8z + 2yz)$.

Factor out the GCF from the first group $(5y^2 – 20y)$:

$5y(y - 4)$.

Factor out the GCF from the second group $(-8z + 2yz)$. The GCF of $-8z$ and $2yz$ is $2z$. Let's factor out $2z$.

$2z(\frac{-8z}{2z} + \frac{2yz}{2z}) = 2z(-4 + y)$, which is $2z(y - 4)$.

The expression becomes $5y(y - 4) + 2z(y - 4)$.

Notice the common binomial factor $(y - 4)$. Factor it out:

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

The factored form is $(y - 4)(5y + 2z)$.

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(viii) Factorise 10ab + 4a + 5b + 2.

This is a four-term polynomial. Let's try grouping.

Group the first two terms and the last two terms:

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

Factor out the GCF from the first group $(10ab + 4a)$:

$2a(5b + 2)$.

The second group $(5b + 2)$ can be written as $1(5b + 2)$.

The expression becomes $2a(5b + 2) + 1(5b + 2)$.

Notice the common binomial factor $(5b + 2)$. Factor it out:

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

The factored form is $(5b + 2)(2a + 1)$.

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(ix) Factorise 6xy – 4y + 6 – 9x.

This is a four-term polynomial. Let's try grouping.

Group the first two terms and the last two terms:

$(6xy – 4y) + (6 – 9x)$.

Factor out the GCF from the first group $(6xy – 4y)$:

$2y(3x - 2)$.

Factor out the GCF from the second group $(6 – 9x)$. The GCF is 3.

$3(2 - 3x)$.

The expression becomes $2y(3x - 2) + 3(2 - 3x)$.

Notice that $(2 - 3x)$ is the negative of $(3x - 2)$. We can write $(2 - 3x) = -(3x - 2)$.

Substitute this into the expression:

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

Notice the common binomial factor $(3x - 2)$. Factor it out:

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

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

We will use the identity for the difference of squares: $A^2 - B^2 = (A + B)(A - B)$. We may also use the perfect square identities: $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$.

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(i) Factorise $a^4 – b^4$.

We can write $a^4$ as $(a^2)^2$ and $b^4$ as $(b^2)^2$.

So, $a^4 – b^4 = (a^2)^2 – (b^2)^2$.

This is a difference of squares with $A = a^2$ and $B = b^2$.

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

The factor $(a^2 - b^2)$ is itself a difference of squares ($a^2 - b^2$).

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

Substitute this back into the expression:

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

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

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(ii) Factorise $p^4 – 81$.

We can write $p^4$ as $(p^2)^2$ and 81 as $9^2$.

So, $p^4 – 81 = (p^2)^2 – 9^2$.

This is a difference of squares with $A = p^2$ and $B = 9$.

$(p^2)^2 – 9^2 = (p^2 + 9)(p^2 - 9)$.

The factor $(p^2 - 9)$ is itself a difference of squares ($p^2 - 3^2$).

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

Substitute this back into the expression:

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

The factored form is $(p^2 + 9)(p + 3)(p - 3)$.

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(iii) Factorise $x^4 – (y + z)^4$.

We can write $x^4$ as $(x^2)^2$ and $(y + z)^4$ as $((y + z)^2)^2$.

So, $x^4 – (y + z)^4 = (x^2)^2 – ((y + z)^2)^2$.

This is a difference of squares with $A = x^2$ and $B = (y + z)^2$.

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

Now, consider the second factor $(x^2 - (y + z)^2)$. This is also a difference of squares, $C^2 - D^2$, where $C = x$ and $D = (y + z)$.

$x^2 - (y + z)^2 = (x + (y + z))(x - (y + z))$.

$x^2 - (y + z)^2 = (x + y + z)(x - y - z)$.

The first factor $(x^2 + (y + z)^2)$ can be expanded but cannot be factored further using real numbers as a sum of squares.

Substitute the factored second term back into the expression:

$(x^2 + (y + z)^2)(x^2 - (y + z)^2) = (x^2 + (y + z)^2)(x + y + z)(x - y - z)$.

The term $(y+z)^2$ can be expanded: $(y+z)^2 = y^2 + 2yz + z^2$.

So, $x^2 + (y + z)^2 = x^2 + y^2 + 2yz + z^2$.

The factored form is $(x^2 + y^2 + 2yz + z^2)(x + y + z)(x - y - z)$.

The fully factored form is:

$(x^2 + (y+z)^2)(x + y + z)(x - y - z)$

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(iv) Factorise $x^4 – (x – z)^4$.

We can write $x^4$ as $(x^2)^2$ and $(x – z)^4$ as $((x – z)^2)^2$.

So, $x^4 – (x – z)^4 = (x^2)^2 – ((x – z)^2)^2$.

This is a difference of squares with $A = x^2$ and $B = (x – z)^2$.

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

Now, consider the second factor $(x^2 - (x – z)^2)$. This is also a difference of squares, $C^2 - D^2$, where $C = x$ and $D = (x – z)$.

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

Simplify each part:

First part: $x + (x – z) = x + x - z = 2x - z$.

Second part: $x - (x – z) = x - x + z = z$.

So, $(x^2 - (x – z)^2) = (2x - z)(z)$.

Now consider the first factor $(x^2 + (x – z)^2)$. Expand $(x-z)^2 = x^2 - 2xz + z^2$.

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

Combine the factored second term and the simplified first term:

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

The factored form is $z(2x - z)(2x^2 - 2xz + z^2)$.

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(v) Factorise $a^4 – 2a^2b^2 + b^4$.

This expression looks like a perfect square trinomial.

We can write $a^4$ as $(a^2)^2$ and $b^4$ as $(b^2)^2$.

The middle term $-2a^2b^2$ is $-2 \times (a^2) \times (b^2)$.

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

This matches the form $X^2 - 2XY + Y^2 = (X - Y)^2$, where $X = a^2$ and $Y = b^2$.

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

The factor $(a^2 - b^2)$ is a difference of squares, $a^2 - b^2 = (a + b)(a - b)$.

Substitute this into the expression:

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

The factored form is $(a^2 - b^2)^2$ or $(a + b)^2 (a - b)^2$.

To factorise quadratic trinomials of the form $x^2 + bx + c$, we look for two numbers $p$ and $q$ such that $p + q = b$ and $p \times q = c$. The factored form will then be $(x + p)(x + q)$.

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

Here, $b = 6$ and $c = 8$. We need two numbers whose sum is 6 and product is 8.

Pairs of integers whose product is 8: (1, 8), (-1, -8), (2, 4), (-2, -4).

Check their sums:

$1 + 8 = 9$

$-1 + (-8) = -9$

$2 + 4 = 6$. This pair works.

$-2 + (-4) = -6$.

The two numbers are 2 and 4.

Rewrite the middle term $6p$ as $2p + 4p$:

$p^2 + 6p + 8 = p^2 + 2p + 4p + 8$.

Factor by grouping:

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

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

$(p + 2)(p + 4)$.

The factored form is $(p + 2)(p + 4)$.

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(ii) Factorise $q^2 – 10q + 21$.

Here, $b = -10$ and $c = 21$. We need two numbers whose sum is -10 and product is 21.

Since the product is positive and the sum is negative, both numbers must be negative.

Pairs of negative integers whose product is 21: (-1, -21), (-3, -7).

Check their sums:

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

$-3 + (-7) = -10$. This pair works.

The two numbers are -3 and -7.

Rewrite the middle term $-10q$ as $-3q - 7q$:

$q^2 – 10q + 21 = q^2 - 3q - 7q + 21$.

Factor by grouping:

$(q^2 - 3q) + (-7q + 21) = q(q - 3) - 7(q - 3)$.

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

$(q - 3)(q - 7)$.

The factored form is $(q - 3)(q - 7)$.

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(iii) Factorise $p^2 + 6p – 16$.

Here, $b = 6$ and $c = -16$. We need two numbers whose sum is 6 and product is -16.

Since the product is negative, one number must be positive and the other negative. The positive number must be larger than the negative number since the sum is positive.

Pairs of integers whose product is -16: (1, -16), (-1, 16), (2, -8), (-2, 8), (4, -4).

Check their sums:

$1 + (-16) = -15$.

$-1 + 16 = 15$.

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

$-2 + 8 = 6$. This pair works.

$4 + (-4) = 0$.

The two numbers are -2 and 8.

Rewrite the middle term $6p$ as $-2p + 8p$:

$p^2 + 6p – 16 = p^2 - 2p + 8p - 16$.

Factor by grouping:

$(p^2 - 2p) + (8p - 16) = p(p - 2) + 8(p - 2)$.

Factor out the common binomial factor $(p - 2)$:

$(p - 2)(p + 8)$.

The factored form is $(p - 2)(p + 8)$.

(i) We need to perform the division $-20x^4 \div 10x^2$.

We can write this as a fraction:

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

Divide the numerical coefficients and the variable parts separately:

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

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

Using the law of exponents $\frac{a^m}{a^n} = a^{m-n}$:

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

Multiply the results:

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

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

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(ii) We need to perform the division $7x^2y^2z^2 \div 14xyz$.

We can write this as a fraction:

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

Divide the numerical coefficients and the variable parts separately:

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

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

Using the law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for each variable:

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

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

$\frac{z^2}{z} = z^{2-1} = z^1 = z$.

Multiply the results:

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

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

We need to divide the expression $24(x^2yz + xy^2z + xyz^2)$ by $8xyz$.

The division can be written as:

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

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Method 1: Direct Division

In this method, we distribute the constant 24 to each term inside the parenthesis in the numerator, and then divide each term of the resulting polynomial by the monomial denominator.

The expression becomes:

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

Divide each term in the numerator by the denominator $8xyz$:

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

Perform each division by dividing the numerical coefficients and the variable parts:

For the first term: $\frac{24}{8} \times \frac{x^2}{x} \times \frac{y}{y} \times \frac{z}{z} = 3 \times x^{2-1} \times y^{1-1} \times z^{1-1} = 3x$.

For the second term: $\frac{24}{8} \times \frac{x}{x} \times \frac{y^2}{y} \times \frac{z}{z} = 3 \times x^{1-1} \times y^{2-1} \times z^{1-1} = 3y$.

For the third term: $\frac{24}{8} \times \frac{x}{x} \times \frac{y}{y} \times \frac{z^2}{z} = 3 \times x^{1-1} \times y^{1-1} \times z^{2-1} = 3z$.

Combine the results:

$3x + 3y + 3z$.

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Method 2: Factorisation

In this method, we factorise the numerator and then cancel out any common factors with the denominator.

Consider the numerator: $24(x^2yz + xy^2z + xyz^2)$.

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

Find the greatest common factor (GCF) of the terms $x^2yz$, $xy^2z$, and $xyz^2$.

The common variables are $x, y, z$. The lowest power of $x$ is $x^1$, the lowest power of $y$ is $y^1$, and the lowest power of $z$ is $z^1$.

The GCF is $xyz$.

Factor out $xyz$ from the expression inside the parenthesis:

$x^2yz + xy^2z + xyz^2 = xyz(\frac{x^2yz}{xyz} + \frac{xy^2z}{xyz} + \frac{xyz^2}{xyz}) = xyz(x + y + z)$.

Substitute this factored form back into the numerator:

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

Now, perform the division:

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

Cancel the common factors in the numerator and the denominator. The numerical factors 24 and 8 have a common factor of 8. The variables $xyz$ are common in both numerator and denominator.

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

This simplifies to:

$3(x + y + z)$

Distribute the 3:

$3x + 3y + 3z$.

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Both methods give the same result.

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

We need to divide $44(x^4 – 5x^3 – 24x^2)$ by $11x (x – 8)$.

We can write this division as:

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

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Let's factorise the numerator.

Numerator = $44(x^4 – 5x^3 – 24x^2)$.

Consider the polynomial inside the parenthesis: $x^4 – 5x^3 – 24x^2$.

We can factor out the greatest common factor (GCF) from these terms. The GCF of $x^4$, $x^3$, and $x^2$ is $x^2$.

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

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Now, factorise the quadratic trinomial $x^2 – 5x – 24$.

We look for two numbers whose product is $-24$ and whose sum is $-5$.

The numbers are $3$ and $-8$ (since $3 \times (-8) = -24$ and $3 + (-8) = -5$).

So, $x^2 – 5x – 24 = (x + 3)(x - 8)$.

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Substitute this factored form back into the numerator:

Numerator = $44x^2(x + 3)(x - 8)$.

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Now perform the division using the factored numerator and the given denominator:

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

Cancel out the common factors in the numerator and the denominator:

Divide the numerical coefficients: $\frac{44}{11} = 4$.

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

Cancel the binomial factor $(x - 8)$ (assuming $x \neq 8$).

The expression simplifies to:

$4x(x + 3)$

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

We need to divide the expression $z(5z^2 – 80)$ by $5z(z + 4)$.

We can write this division as a fraction:

$\frac{z(5z^2 – 80)}{5z(z + 4)}$

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Let's factorise the numerator: $z(5z^2 – 80)$.

Consider the expression inside the parenthesis: $5z^2 – 80$.

We can factor out the common numerical factor 5:

$5z^2 – 80 = 5(z^2 – 16)$.

Now, consider the expression inside the new parenthesis: $z^2 – 16$. This is a difference of two perfect squares, $z^2 - 4^2$.

Using the identity $a^2 - b^2 = (a+b)(a-b)$, we factorise $z^2 – 16$:

$z^2 – 16 = (z + 4)(z - 4)$.

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

$5(z^2 – 16) = 5(z + 4)(z - 4)$.

Now, substitute this factored form back into the numerator:

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

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Now perform the division using the factored numerator and the given denominator:

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

Cancel the common factors in the numerator and the denominator. The common factors are $5$, $z$, and $(z+4)$ (assuming $z \neq 0$ and $z \neq -4$).

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

The remaining factor in the numerator is $(z - 4)$.

The expression simplifies to:

$z - 4$.

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

(i) We need to divide $28x^4$ by $56x$.

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

Simplify the numerical part: $\frac{28}{56} = \frac{1}{2}$.

Simplify the variable part using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$: $\frac{x^4}{x} = x^{4-1} = x^3$.

Combine the results: $\frac{1}{2} \times x^3 = \frac{1}{2}x^3$.

The result is $\frac{1}{2}x^3$.

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(ii) We need to divide $-36y^3$ by $9y^2$.

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

Simplify the numerical part: $\frac{-36}{9} = -4$.

Simplify the variable part: $\frac{y^3}{y^2} = y^{3-2} = y^1 = y$.

Combine the results: $-4 \times y = -4y$.

The result is $-4y$.

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(iii) We need to divide $66pq^2r^3$ by $11qr^2$.

$\frac{66pq^2r^3}{11qr^2} = \frac{66}{11} \times \frac{p}{1} \times \frac{q^2}{q} \times \frac{r^3}{r^2}$

Simplify the numerical part: $\frac{66}{11} = 6$.

Simplify the variable parts: $\frac{p}{1} = p$, $\frac{q^2}{q} = q^{2-1} = q$, $\frac{r^3}{r^2} = r^{3-2} = r$.

Combine the results: $6 \times p \times q \times r = 6pqr$.

The result is $6pqr$.

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(iv) We need to divide $34x^3y^3z^3$ by $51xy^2z^3$.

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

Simplify the numerical part: $\frac{34}{51} = \frac{34 \div 17}{51 \div 17} = \frac{2}{3}$.

Simplify the variable parts: $\frac{x^3}{x} = x^{3-1} = x^2$, $\frac{y^3}{y^2} = y^{3-2} = y$, $\frac{z^3}{z^3} = z^{3-3} = z^0 = 1$.

Combine the results: $\frac{2}{3} \times x^2 \times y \times 1 = \frac{2}{3}x^2y$.

The result is $\frac{2}{3}x^2y$.

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(v) We need to divide $12a^8b^8$ by $– 6a^6b^4$.

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

Simplify the numerical part: $\frac{12}{-6} = -2$.

Simplify the variable parts: $\frac{a^8}{a^6} = a^{8-6} = a^2$, $\frac{b^8}{b^4} = b^{8-4} = b^4$.

Combine the results: $-2 \times a^2 \times b^4 = -2a^2b^4$.

The result is $-2a^2b^4$.

To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial separately and then add or subtract the results.

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(i) Divide $(5x^2 – 6x)$ by $3x$.

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

Divide the first term: $\frac{5x^2}{3x} = \frac{5}{3} x^{2-1} = \frac{5}{3}x$.

Divide the second term: $\frac{6x}{3x} = \frac{6}{3} x^{1-1} = 2 x^0 = 2 \times 1 = 2$.

Combine the results: $\frac{5}{3}x - 2$.

The result is $\frac{5}{3}x - 2$.

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(ii) Divide $(3y^8 – 4y^6 + 5y^4)$ by $y^4$.

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

Divide the first term: $\frac{3y^8}{y^4} = 3y^{8-4} = 3y^4$.

Divide the second term: $\frac{4y^6}{y^4} = 4y^{6-4} = 4y^2$.

Divide the third term: $\frac{5y^4}{y^4} = 5y^{4-4} = 5y^0 = 5$.

Combine the results: $3y^4 - 4y^2 + 5$.

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

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(iii) Divide $8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)$ by $4x^2y^2z^2$.

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

We can divide the numerical coefficients first: $\frac{8}{4} = 2$.

The expression becomes $2 \times \frac{x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3}{x^2y^2z^2}$.

Now, divide each term in the numerator by the denominator $x^2y^2z^2$:

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

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

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

Combine the results inside the parenthesis: $x + y + z$.

Multiply by the initial numerical factor 2: $2(x + y + z)$.

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

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(iv) Divide $(x^3 + 2x^2 +3x)$ by $2x$.

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

Divide the first term: $\frac{x^3}{2x} = \frac{1}{2} x^{3-1} = \frac{1}{2}x^2$.

Divide the second term: $\frac{2x^2}{2x} = \frac{2}{2} x^{2-1} = 1 x^1 = x$.

Divide the third term: $\frac{3x}{2x} = \frac{3}{2} x^{1-1} = \frac{3}{2} x^0 = \frac{3}{2}$.

Combine the results: $\frac{1}{2}x^2 + x + \frac{3}{2}$.

The result is $\frac{1}{2}x^2 + x + \frac{3}{2}$.

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(v) Divide $(p^3q^6 – p^6q^3)$ by $p^3q^3$.

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

Divide the first term: $\frac{p^3q^6}{p^3q^3} = p^{3-3}q^{6-3} = p^0q^3 = 1 \times q^3 = q^3$.

Divide the second term: $\frac{p^6q^3}{p^3q^3} = p^{6-3}q^{3-3} = p^3q^0 = p^3 \times 1 = p^3$.

Combine the results: $q^3 - p^3$.

The result is $q^3 - p^3$.

(i) $(10x - 25) \div 5$

We can write the expression as $\frac{10x - 25}{5}$.

Factor out the common term from the numerator: $10x - 25 = 5(2x - 5)$.

So, the expression becomes $\frac{5(2x - 5)}{5}$.

Now, we can cancel out the common factor: $\frac{\cancel{5}(2x - 5)}{\cancel{5}}$

The result is $2x - 5$.

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(ii) $(10x - 25) \div (2x - 5)$

We can write the expression as $\frac{10x - 25}{2x - 5}$.

Factor out the common term from the numerator: $10x - 25 = 5(2x - 5)$.

So, the expression becomes $\frac{5(2x - 5)}{2x - 5}$.

Now, we can cancel out the common factor $(2x - 5)$: $\frac{5\cancel{(2x - 5)}}{\cancel{2x - 5}}$

The result is $5$.

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

We can write the expression as $\frac{10y(6y + 21)}{5(2y + 7)}$.

Factor out the common term from the bracket in the numerator: $6y + 21 = 3(2y + 7)$.

So, the expression becomes $\frac{10y \times 3(2y + 7)}{5(2y + 7)}$.

Now, we can rearrange and cancel out the common factors: $\frac{\cancel{10}^{2}y \times 3 \cancel{(2y + 7)}}{\cancel{5} \cancel{(2y + 7)}}$

The result is $2y \times 3 = 6y$.

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

We can write the expression as $\frac{9x^{2}y^{2}(3z - 24)}{27xy(z - 8)}$.

Factor out the common term from the bracket in the numerator: $3z - 24 = 3(z - 8)$.

So, the expression becomes $\frac{9x^{2}y^{2} \times 3(z - 8)}{27xy(z - 8)}$.

Now, we can rearrange and cancel out the common factors: $\frac{\cancel{9}x^{\cancel{2}}y^{\cancel{2}} \times \cancel{3}\cancel{(z - 8)}}{\cancel{27}_{3} \cancel{x}\cancel{y}\cancel{(z - 8)}}$

Simplify the remaining terms: $\frac{\cancel{3}xy}{\cancel{3}} = xy$.

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

Factor out common terms from the brackets in the numerator: $3a - 12 = 3(a - 4)$ $5b - 30 = 5(b - 6)$

So, the expression becomes $\frac{96abc \times 3(a - 4) \times 5(b - 6)}{144(a - 4)(b - 6)}$.

Now, we can rearrange and cancel out the common factors: $\frac{\cancel{96}^{2}abc \times \cancel{3}\cancel{(a - 4)} \times \cancel{5}\cancel{(b - 6)}}{\cancel{144}_{3} \cancel{(a - 4)}\cancel{(b - 6)}}$

Simplify the remaining terms: $\frac{2 \times 5 \times abc}{3} = \frac{10abc}{3}$. (Correction in calculation, $96 \times 3 = 288$, $144 \times 2 = 288$. So $\frac{96 \times 3}{144} = 2$)

Let's redo the coefficient cancellation: $\frac{96 \times 3 \times 5}{144}$. $96 \times 3 = 288$. So, $\frac{288 \times 5}{144}$. Since $288 = 2 \times 144$, this simplifies to $2 \times 5 = 10$.

$\frac{\cancel{96}^{2}abc \times \cancel{3}^{1}\cancel{(a - 4)} \times 5\cancel{(b - 6)}}{\cancel{144}_{1} \cancel{(a - 4)}\cancel{(b - 6)}}$ (Using $96 \times 3 / 144 = 288 / 144 = 2$)

The result is $2 \times 5 \times abc = 10abc$.

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

We can write the expression as $\frac{5(2x + 1) (3x + 5)}{2x + 1}$.

Now, we can cancel out the common factor $(2x + 1)$: $\frac{5\cancel{(2x + 1)} (3x + 5)}{\cancel{2x + 1}}$

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

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

We can write the expression as $\frac{26xy(x + 5)(y – 4)}{13x(y – 4)}$.

Now, we can cancel out the common factors: $\frac{\cancel{26}^{2}\cancel{x}y(x + 5)\cancel{(y – 4)}}{\cancel{13}\cancel{x}\cancel{(y – 4)}}$

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

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

We can write the expression as $\frac{52pqr(p + q) (q + r) (r + p)}{104pq(q + r) (r + p)}$.

Now, we can cancel out the common factors: $\frac{\cancel{52}^{1}\cancel{p}\cancel{q}r(p + q) \cancel{(q + r)} \cancel{(r + p)}}{\cancel{104}_{2}\cancel{p}\cancel{q}\cancel{(q + r)} \cancel{(r + p)}}$

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

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

We can write the expression as $\frac{20(y + 4) (y^{2} + 5y + 3)}{5(y + 4)}$.

Now, we can cancel out the common factors: $\frac{\cancel{20}^{4}\cancel{(y + 4)} (y^{2} + 5y + 3)}{\cancel{5}\cancel{(y + 4)}}$

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

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

We can write the expression as $\frac{x(x + 1) (x + 2) (x + 3)}{x(x + 1)}$.

Now, we can cancel out the common factors: $\frac{\cancel{x}\cancel{(x + 1)} (x + 2) (x + 3)}{\cancel{x}\cancel{(x + 1)}}$

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

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

First, factorise the quadratic expression $y^{2} + 7y + 10$. We look for two numbers that multiply to $10$ and add up to $7$. These numbers are $2$ and $5$.

So, $y^{2} + 7y + 10 = (y + 2)(y + 5)$.

Now, perform the division: $\frac{(y + 2)(y + 5)}{(y + 5)}$.

Cancel the common factor $(y + 5)$: $\frac{(y + 2)\cancel{(y + 5)}}{\cancel{(y + 5)}}$

The result is $y + 2$.

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(ii) $(m^{2} – 14m – 32) \div (m + 2)$

First, factorise the quadratic expression $m^{2} – 14m – 32$. We look for two numbers that multiply to $-32$ and add up to $-14$. These numbers are $-16$ and $2$.

So, $m^{2} – 14m – 32 = (m - 16)(m + 2)$.

Now, perform the division: $\frac{(m - 16)(m + 2)}{(m + 2)}$.

Cancel the common factor $(m + 2)$: $\frac{(m - 16)\cancel{(m + 2)}}{\cancel{(m + 2)}}$

The result is $m - 16$.

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

First, factorise the expression $5p^{2} – 25p + 20$. Take out the common factor $5$ from all terms: $5(p^{2} – 5p + 4)$.

Now, factorise the quadratic expression $p^{2} – 5p + 4$. We look for two numbers that multiply to $4$ and add up to $-5$. These numbers are $-1$ and $-4$.

So, $p^{2} – 5p + 4 = (p - 1)(p - 4)$.

Thus, $5p^{2} – 25p + 20 = 5(p - 1)(p - 4)$.

Now, perform the division: $\frac{5(p - 1)(p - 4)}{(p - 1)}$.

Cancel the common factor $(p - 1)$: $\frac{5\cancel{(p - 1)}(p - 4)}{\cancel{(p - 1)}}$

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

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

First, factorise the quadratic expression $z^{2} + 6z – 16$. We look for two numbers that multiply to $-16$ and add up to $6$. These numbers are $8$ and $-2$.

So, $z^{2} + 6z – 16 = (z + 8)(z - 2)$.

Now, perform the division: $\frac{4yz(z + 8)(z - 2)}{2y(z + 8)}$.

Cancel the common factors: $\frac{\cancel{4}^{2}\cancel{y}z\cancel{(z + 8)}(z - 2)}{\cancel{2}\cancel{y}\cancel{(z + 8)}}$

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

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(v) $5pq(p^{2} – q^{2}) \div 2p(p + q)$

First, factorise the expression $p^{2} – q^{2}$. This is a difference of squares, which factors as $(p - q)(p + q)$.

So, $p^{2} – q^{2} = (p - q)(p + q)$.

Now, perform the division: $\frac{5pq(p - q)(p + q)}{2p(p + q)}$.

Cancel the common factors: $\frac{5\cancel{p}q(p - q)\cancel{(p + q)}}{2\cancel{p}\cancel{(p + q)}}$

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

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

First, factorise the expression $9x^{2} – 16y^{2}$. This is a difference of squares, which factors as $(3x)^{2} – (4y)^{2} = (3x - 4y)(3x + 4y)$.

So, $9x^{2} – 16y^{2} = (3x - 4y)(3x + 4y)$.

Now, perform the division: $\frac{12xy(3x - 4y)(3x + 4y)}{4xy(3x + 4y)}$.

Cancel the common factors: $\frac{\cancel{12}^{3}\cancel{x}\cancel{y}(3x - 4y)\cancel{(3x + 4y)}}{\cancel{4}\cancel{x}\cancel{y}\cancel{(3x + 4y)}}$

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

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

First, factorise the expression $50y^{2} – 98$. Take out the common factor $2$: $2(25y^{2} – 49)$.

Now, factorise the expression $25y^{2} – 49$. This is a difference of squares, which factors as $(5y)^{2} – 7^{2} = (5y - 7)(5y + 7)$.

So, $50y^{2} – 98 = 2(5y - 7)(5y + 7)$.

The numerator is $39y^{3} \times 2(5y - 7)(5y + 7) = 78y^{3}(5y - 7)(5y + 7)$.

Now, perform the division: $\frac{78y^{3}(5y - 7)(5y + 7)}{26y^{2}(5y + 7)}$.

Cancel the common factors (coefficients, powers of $y$, and the bracket term): $\frac{\cancel{78}^{3}y^{\cancel{3}}(5y - 7)\cancel{(5y + 7)}}{\cancel{26}\cancel{y^{2}}\cancel{(5y + 7)}}$

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

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The given equation is $4(x – 5) = 4x – 5$.

Let's evaluate the Left Hand Side (LHS): LHS $= 4(x - 5)$

Using the distributive property of multiplication over subtraction, we multiply $4$ by each term inside the bracket: LHS $= 4 \times x - 4 \times 5$

LHS $= 4x - 20$

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The Right Hand Side (RHS) is $4x - 5$.

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Comparing the LHS and RHS: $4x - 20$ and $4x - 5$

Since $-20 \neq -5$, the LHS is not equal to the RHS.

Therefore, the given equation is incorrect.

The error is in applying the distributive property. The number $4$ should be multiplied by both $x$ and $5$, not just $x$.

The correct equation would be $4(x - 5) = 4x - 20$.

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The given equation is $x(3x + 2) = 3x^2 + 2$.

Let's evaluate the Left Hand Side (LHS): LHS $= x(3x + 2)$

Using the distributive property of multiplication over addition, we multiply $x$ by each term inside the bracket: LHS $= x \times 3x + x \times 2$

LHS $= 3x^2 + 2x$

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The Right Hand Side (RHS) is $3x^2 + 2$.

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Comparing the LHS and RHS: $3x^2 + 2x$ and $3x^2 + 2$

Since $2x \neq 2$ (unless $x=1$), the LHS is generally not equal to the RHS.

Therefore, the given equation is incorrect.

The error is in applying the distributive property. The term $x$ should be multiplied by both $3x$ and $2$, not just $3x$.

The correct expansion of $x(3x + 2)$ is $3x^2 + 2x$.

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The given equation is $2x + 3y = 5xy$.

Let's look at the Left Hand Side (LHS): $2x + 3y$.

The terms $2x$ and $3y$ are unlike terms because they have different variables ($x$ and $y$).

Unlike terms cannot be combined by addition or subtraction. The expression $2x + 3y$ is already in its simplest form and cannot be written as a single term.

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The Right Hand Side (RHS) is $5xy$.

This is a single term representing the product of $5$, $x$, and $y$.

---

Comparing the LHS and RHS, we see that a sum of unlike terms ($2x + 3y$) is being equated to a product term ($5xy$). These are fundamentally different types of expressions.

For example, if we take $x=1$ and $y=1$: LHS $= 2(1) + 3(1) = 2 + 3 = 5$. RHS $= 5(1)(1) = 5$. In this specific case, the equation holds true.

However, if we take $x=2$ and $y=3$: LHS $= 2(2) + 3(3) = 4 + 9 = 13$. RHS $= 5(2)(3) = 5 \times 6 = 30$. Here, $13 \neq 30$.

The equation $2x + 3y = 5xy$ is not an identity; it does not hold true for all values of $x$ and $y$. The error is in assuming that $2x$ and $3y$ can be combined through addition into $5xy$.

The sum of unlike terms $2x + 3y$ cannot be simplified to $5xy$.

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The given equation is $x + 2x + 3x = 5x$.

Let's evaluate the Left Hand Side (LHS): LHS $= x + 2x + 3x$

These are like terms because they all have the variable $x$ raised to the power of $1$.

We can combine the coefficients of these like terms: LHS $= (1 + 2 + 3)x$

LHS $= 6x$

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The Right Hand Side (RHS) is $5x$.

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Comparing the LHS and RHS: $6x$ and $5x$

The equation $6x = 5x$ is only true if $x=0$. It is not true for all values of $x$.

Therefore, the given equation is incorrect as an identity.

The error is in incorrectly adding the coefficients of the like terms on the LHS. The sum of $1$, $2$, and $3$ is $6$, not $5$.

The correct simplification of the LHS is $6x$.

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The given equation is $5y + 2y + y – 7y = 0$.

Let's evaluate the Left Hand Side (LHS): LHS $= 5y + 2y + y – 7y$

These are all like terms because they all have the variable $y$ raised to the power of $1$. The term $y$ can be written as $1y$.

We can combine the coefficients of these like terms: LHS $= (5 + 2 + 1 – 7)y$

First, add the positive coefficients: $5 + 2 + 1 = 8$.

So, the expression becomes $(8 – 7)y$.

Perform the subtraction: $8 – 7 = 1$.

LHS $= 1y = y$

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The Right Hand Side (RHS) is $0$.

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Comparing the LHS and RHS: $y$ and $0$

The equation $y = 0$ is only true if the variable $y$ has the value $0$. It is not true for all values of $y$.

Therefore, the given equation is incorrect as an identity.

The error is in equating the simplified expression $y$ to $0$, implying that the sum of terms was expected to be $0$ always, which is only the case when $y=0$.

The correct simplification of the LHS is $y$.

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The given equation is $3x + 2x = 5x^2$.

Let's evaluate the Left Hand Side (LHS): LHS $= 3x + 2x$

These are like terms because they both have the variable $x$ raised to the power of $1$.

To add like terms, we add their coefficients and keep the variable part the same: LHS $= (3 + 2)x$

LHS $= 5x$

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The Right Hand Side (RHS) is $5x^2$.

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Comparing the LHS and RHS: $5x$ and $5x^2$

The terms $5x$ and $5x^2$ are unlike terms because the variable $x$ has different powers ($1$ in $5x$ and $2$ in $5x^2$). They are not equal in general (they are equal only if $x=0$ or $x=1$).

Therefore, the given equation is incorrect as an identity.

The error is in adding the like terms on the LHS. When adding like terms, the variable part (including its power) does not change. Only the coefficients are added. The variable part $x$ should remain $x$, not become $x^2$.

The correct simplification of the LHS is $5x$.

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The given equation is $(2x)^2 + 4(2x) + 7 = 2x^2 + 8x + 7$.

Let's evaluate the Left Hand Side (LHS): LHS $= (2x)^2 + 4(2x) + 7$

Evaluate the first term $(2x)^2$. The square of a product is the product of the squares: $(2x)^2 = 2^2 \times x^2 = 4x^2$.

Evaluate the second term $4(2x)$. Multiply the coefficients: $4(2x) = 4 \times 2 \times x = 8x$.

Substitute these back into the LHS: LHS $= 4x^2 + 8x + 7$

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The Right Hand Side (RHS) is $2x^2 + 8x + 7$.

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Comparing the LHS and RHS: $4x^2 + 8x + 7$ and $2x^2 + 8x + 7$

These expressions are not equal because the coefficients of the $x^2$ term are different ($4$ on the LHS and $2$ on the RHS).

Therefore, the given equation is incorrect as an identity.

The error is in calculating $(2x)^2$. $(2x)^2$ means $(2x) \times (2x)$, which is $4x^2$, not $2x^2$. The coefficient inside the bracket must also be squared.

The correct expansion of the LHS is $4x^2 + 8x + 7$.

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The given chain of equality is $(2x)^2 + 5x = 4x + 5x = 9x$. This statement implies that $(2x)^2 + 5x$ is equal to $4x + 5x$, and $4x + 5x$ is equal to $9x$.

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Let's evaluate the first expression: $(2x)^2 + 5x$. We need to calculate $(2x)^2$ first.

$(2x)^2 = (2x) \times (2x) = 2 \times 2 \times x \times x = 4x^2$.

So, the first expression is $4x^2 + 5x$.

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Now let's evaluate the second expression: $4x + 5x$. These are like terms. We can combine their coefficients: $4x + 5x = (4 + 5)x = 9x$.

So, the second expression is $9x$.

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The third expression is $9x$, which is already simplified.

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The given statement $4x^2 + 5x = 9x = 9x$ simplifies to $4x^2 + 5x = 9x$.

Let's check if $4x^2 + 5x = 9x$ is true for all values of $x$. Subtract $5x$ from both sides: $4x^2 + 5x - 5x = 9x - 5x$ $4x^2 = 4x$

Divide both sides by $4$: $x^2 = x$

Rearrange the equation: $x^2 - x = 0$

Factor out $x$: $x(x - 1) = 0$

This equation is only true if $x = 0$ or $x - 1 = 0$ (i.e., $x = 1$). It is not true for all values of $x$.

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The error occurs in the first part of the equality: $(2x)^2 + 5x = 4x + 5x$.

The term $(2x)^2$ was incorrectly evaluated or treated as $4x$ or something similar that would allow it to be combined with $5x$ to get $9x$.

The correct evaluation of $(2x)^2$ is $4x^2$.

The correct simplification of the first expression is $(2x)^2 + 5x = 4x^2 + 5x$.

The correct simplification of $4x + 5x$ is $9x$.

So, the correct statement should be $4x^2 + 5x = 4x + 5x$ (which is false) or $4x^2 + 5x = 9x$ (which is false for most $x$), or $4x + 5x = 9x$ (which is true). The original statement incorrectly equates the first expression to the subsequent ones.

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The given equation is $(3x + 2)^2 = 3x^2 + 6x + 4$.

Let's evaluate the Left Hand Side (LHS): LHS $= (3x + 2)^2$

We use the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = 3x$ and $b = 2$.

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

Evaluate each term: $(3x)^2 = 3^2 \times x^2 = 9x^2$. $2(3x)(2) = 2 \times 3 \times x \times 2 = 12x$. $(2)^2 = 4$.

Substitute these values back into the expression: LHS $= 9x^2 + 12x + 4$

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The Right Hand Side (RHS) is $3x^2 + 6x + 4$.

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Comparing the LHS and RHS: $9x^2 + 12x + 4$ and $3x^2 + 6x + 4$

These expressions are not equal because the coefficients of the $x^2$ term ($9$ vs $3$) and the coefficients of the $x$ term ($12$ vs $6$) are different.

Therefore, the given equation is incorrect as an identity.

The error is in the expansion of $(3x + 2)^2$. The coefficient $3$ in $3x$ must be squared when calculating $(3x)^2$, resulting in $9x^2$, not $3x^2$. Also, the middle term $2ab$ was calculated incorrectly or the coefficients were summed instead of multiplied with $2$.

The correct expansion of $(3x + 2)^2$ is $9x^2 + 12x + 4$.

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We need to check the correctness of the substitution and evaluation for each expression when $x = -3$.

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(a) Expression: $x^{2} + 5x + 4$. Substituting $x = -3$.

$(-3)^{2} + 5(-3) + 4$

Evaluate each term: $(-3)^{2} = (-3) \times (-3) = 9$ $5(-3) = 5 \times (-3) = -15$

Substitute these values back into the expression: $9 + (-15) + 4$

$9 - 15 + 4$

$-6 + 4$

$-2$

The calculation provided is $9 + 2 + 4 = 15$.

The correct result is $-2$.

The statement is incorrect. The error is in evaluating $5(-3)$ as $+2$ instead of $-15$.

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(b) Expression: $x^{2} – 5x + 4$. Substituting $x = -3$.

$(-3)^{2} – 5(-3) + 4$

Evaluate each term: $(-3)^{2} = (-3) \times (-3) = 9$ $-5(-3) = (-5) \times (-3) = 15$

Substitute these values back into the expression: $9 + 15 + 4$

$24 + 4$

$28$

The calculation provided is $9 – 15 + 4 = – 2$.

The correct result is $28$.

The statement is incorrect. The error is in evaluating $-5(-3)$ as $-15$ instead of $+15$ and the subsequent addition/subtraction.

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(c) Expression: $x^{2} + 5x$. Substituting $x = -3$.

$(-3)^{2} + 5(-3)$

Evaluate each term: $(-3)^{2} = (-3) \times (-3) = 9$ $5(-3) = 5 \times (-3) = -15$

Substitute these values back into the expression: $9 + (-15)$

$9 - 15$

$-6$

The calculation provided is $– 9 – 15 = – 24$.

The correct result is $-6$.

The statement is incorrect. The error is in evaluating $(-3)^{2}$ as $-9$ instead of $9$ and the subsequent subtraction.

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The given equation is $(y – 3)^2 = y^2 – 9$.

Let's evaluate the Left Hand Side (LHS): LHS $= (y - 3)^2$

We use the algebraic identity for squaring a binomial difference: $(a - b)^2 = a^2 - 2ab + b^2$. Here, $a = y$ and $b = 3$.

LHS $= (y)^2 - 2(y)(3) + (3)^2$

Evaluate each term: $(y)^2 = y^2$. $2(y)(3) = 6y$. $(3)^2 = 9$.

Substitute these values back into the expression: LHS $= y^2 - 6y + 9$

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The Right Hand Side (RHS) is $y^2 – 9$.

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Comparing the LHS and RHS: $y^2 - 6y + 9$ and $y^2 - 9$

These expressions are not equal because the LHS contains a term with $y$ (specifically $-6y$) and a constant term $+9$, while the RHS has no $y$ term and a constant term $-9$.

Therefore, the given equation is incorrect as an identity.

The error is in the expansion of $(y - 3)^2$. It seems to have incorrectly assumed that $(y - 3)^2 = y^2 - 3^2$, which is only true for the difference of squares $y^2 - 3^2 = (y-3)(y+3)$, not for the square of a difference. The middle term $-2(y)(3) = -6y$ is missing in the RHS of the given equation.

The correct expansion of $(y - 3)^2$ is $y^2 - 6y + 9$.

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The given equation is $(z + 5)^2 = z^2 + 25$.

Let's evaluate the Left Hand Side (LHS): LHS $= (z + 5)^2$

We use the algebraic identity for squaring a binomial sum: $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = z$ and $b = 5$.

LHS $= (z)^2 + 2(z)(5) + (5)^2$

Evaluate each term: $(z)^2 = z^2$. $2(z)(5) = 10z$. $(5)^2 = 25$.

Substitute these values back into the expression: LHS $= z^2 + 10z + 25$

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The Right Hand Side (RHS) is $z^2 + 25$.

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Comparing the LHS and RHS: $z^2 + 10z + 25$ and $z^2 + 25$

These expressions are not equal because the LHS contains a term with $z$ (specifically $+10z$) which is missing in the RHS.

Therefore, the given equation is incorrect as an identity.

The error is in the expansion of $(z + 5)^2$. It seems to have incorrectly assumed that $(z + 5)^2 = z^2 + 5^2$, omitting the middle term $2(z)(5) = 10z$.

The correct expansion of $(z + 5)^2$ is $z^2 + 10z + 25$.

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The given equation is $(2a + 3b)(a – b) = 2a^2 – 3b^2$.

Let's evaluate the Left Hand Side (LHS): LHS $= (2a + 3b)(a - b)$

We multiply the two binomials using the distributive property (FOIL method or term by term multiplication). Multiply the first term of the first binomial by each term of the second binomial: $2a \times (a - b) = 2a \times a - 2a \times b = 2a^2 - 2ab$.

Multiply the second term of the first binomial by each term of the second binomial: $3b \times (a - b) = 3b \times a - 3b \times b = 3ab - 3b^2$.

Now, add these results: LHS $= (2a^2 - 2ab) + (3ab - 3b^2)$ LHS $= 2a^2 - 2ab + 3ab - 3b^2$

Combine the like terms ($-2ab$ and $+3ab$): LHS $= 2a^2 + (-2 + 3)ab - 3b^2$ LHS $= 2a^2 + 1ab - 3b^2$ LHS $= 2a^2 + ab - 3b^2$

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The Right Hand Side (RHS) is $2a^2 – 3b^2$.

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Comparing the LHS and RHS: $2a^2 + ab - 3b^2$ and $2a^2 – 3b^2$

These expressions are not equal because the LHS contains a middle term $+ab$, which is missing in the RHS.

Therefore, the given equation is incorrect as an identity.

The error is in the multiplication of the binomials. The cross-product terms (like $2a \times -b$ and $3b \times a$) were either not calculated correctly or omitted. The resulting middle term $+ab$ was not included in the RHS.

The correct expansion of $(2a + 3b)(a - b)$ is $2a^2 + ab - 3b^2$.

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The given equation is $(a + 4)(a + 2) = a^2 + 8$.

Let's evaluate the Left Hand Side (LHS): LHS $= (a + 4)(a + 2)$

We multiply the two binomials using the distributive property (FOIL method).

Multiply the First terms: $a \times a = a^2$. Multiply the Outer terms: $a \times 2 = 2a$. Multiply the Inner terms: $4 \times a = 4a$. Multiply the Last terms: $4 \times 2 = 8$.

Add these four terms: LHS $= a^2 + 2a + 4a + 8$

Combine the like terms ($2a$ and $4a$): LHS $= a^2 + (2 + 4)a + 8$ LHS $= a^2 + 6a + 8$

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The Right Hand Side (RHS) is $a^2 + 8$.

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Comparing the LHS and RHS: $a^2 + 6a + 8$ and $a^2 + 8$

These expressions are not equal because the LHS contains a term with $a$ (specifically $6a$) which is missing in the RHS.

Therefore, the given equation is incorrect as an identity.

The error is in the multiplication of the binomials. The cross-product terms ($a \times 2$ and $4 \times a$) were omitted, leading to the middle term $6a$ being missed in the RHS. It seems only the product of the first terms ($a \times a = a^2$) and the product of the last terms ($4 \times 2 = 8$) were considered.

The correct expansion of $(a + 4)(a + 2)$ is $a^2 + 6a + 8$.

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The given equation is $(a – 4) (a – 2) = a^2 – 8$.

Let's evaluate the Left Hand Side (LHS): LHS $= (a - 4)(a - 2)$

We multiply the two binomials using the distributive property (FOIL method).

Multiply the First terms: $a \times a = a^2$.

Multiply the Outer terms: $a \times (-2) = -2a$.

Multiply the Inner terms: $(-4) \times a = -4a$.

Multiply the Last terms: $(-4) \times (-2) = 8$.

Add these four terms: LHS $= a^2 - 2a - 4a + 8$

Combine the like terms ($-2a$ and $-4a$): LHS $= a^2 + (-2 - 4)a + 8$ LHS $= a^2 - 6a + 8$

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The Right Hand Side (RHS) is $a^2 – 8$.

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Comparing the LHS and RHS: $a^2 - 6a + 8$ and $a^2 - 8$

These expressions are not equal because the LHS contains a middle term $-6a$, which is missing in the RHS, and the constant term is $+8$ on the LHS compared to $-8$ on the RHS.

Therefore, the given equation is incorrect as an identity.

The error is in the multiplication of the binomials. It seems only the product of the first terms ($a \times a = a^2$) and the product of the last terms ($(-4) \times (-2) = 8$) were calculated, and then incorrectly combined as $a^2 - 8$. The cross-product terms ($a \times -2 = -2a$ and $-4 \times a = -4a$) were omitted, leading to the middle term $-6a$ being missed. Also, the constant term resulting from the product of the last terms should be $+8$, not $-8$.

The correct expansion of $(a - 4)(a - 2)$ is $a^2 - 6a + 8$.

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The given equation is $\frac{3x^{2}}{3x^{2}} = 0$.

Let's evaluate the Left Hand Side (LHS): LHS $= \frac{3x^{2}}{3x^{2}}$

For this expression to be defined, the denominator cannot be zero. So, $3x^2 \neq 0$, which implies $x \neq 0$.

Assuming $x \neq 0$, the numerator and the denominator are the same non-zero expression.

When any non-zero quantity is divided by itself, the result is always $1$.

So, LHS $= \frac{\cancel{3x^{2}}}{\cancel{3x^{2}}} = 1$.

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The Right Hand Side (RHS) is $0$.

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Comparing the LHS and RHS: $1$ and $0$

Since $1 \neq 0$, the LHS is not equal to the RHS.

Therefore, the given equation is incorrect.

The error is in assuming that dividing a term by itself results in $0$. Division of any non-zero term by itself results in $1$.

The correct statement is $\frac{3x^{2}}{3x^{2}} = 1$ (provided $x \neq 0$).

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The given equation is $\frac{3x^{2}\;+\;1}{3x^{2}} = 1 + 1 = 2$.

Let's evaluate the Left Hand Side (LHS): LHS $= \frac{3x^{2}\;+\;1}{3x^{2}}$

When a numerator with multiple terms is divided by a single term denominator, the denominator must divide each term in the numerator.

So, we can rewrite the fraction as a sum of two fractions: LHS $= \frac{3x^{2}}{3x^{2}} + \frac{1}{3x^{2}}$

Assuming $x \neq 0$, we can simplify the first term: $\frac{3x^{2}}{3x^{2}} = \frac{\cancel{3x^{2}}}{\cancel{3x^{2}}} = 1$.

The second term $\frac{1}{3x^{2}}$ cannot be simplified further.

So, the correct simplification of the LHS is: LHS $= 1 + \frac{1}{3x^{2}}$

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The Right Hand Side (RHS) is $1 + 1 = 2$.

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Comparing the LHS and RHS: $1 + \frac{1}{3x^{2}}$ and $2$

These expressions are not equal in general. The equation $1 + \frac{1}{3x^{2}} = 2$ implies $\frac{1}{3x^{2}} = 1$, or $3x^{2} = 1$, which is only true for $x = \pm \frac{1}{\sqrt{3}}$. It is not an identity for all values of $x$ (where $x \neq 0$).

Therefore, the given equation is incorrect.

The error is in incorrectly simplifying the fraction $\frac{3x^{2}\;+\;1}{3x^{2}}$. One cannot simply cancel out the $3x^2$ term in the numerator's first term and the denominator when there are other terms being added or subtracted in the numerator. The denominator must divide every term in the numerator. The incorrect simplification treated $\frac{3x^{2}\;+\;1}{3x^{2}}$ as if it were $\frac{3x^{2}}{3x^{2}} + 1$, which is wrong.

The correct simplification is $1 + \frac{1}{3x^{2}}$.

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The given equation is $\frac{3x}{3x + 2} = \frac{1}{2}$.

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Let's analyze the Left Hand Side (LHS): LHS $= \frac{3x}{3x + 2}$.

The numerator is a single term $3x$. The denominator is a sum of two terms, $3x$ and $2$.

In a fraction, a factor can only be cancelled from the numerator and the denominator if it is a factor of the entire numerator and the entire denominator.

In the denominator $3x + 2$, $3x$ is a term, not a factor of the whole expression $(3x + 2)$ (unless $x=0$, but then the numerator is 0, and the denominator is 2, giving $\frac{0}{2}=0$, not $\frac{1}{2}$). The term $2$ is added to $3x$. Therefore, $3x$ cannot be cancelled out from the numerator and the first term of the denominator.

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The Right Hand Side (RHS) is $\frac{1}{2}$.

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To check if the equation is true, we can cross-multiply: $3x \times 2 = 1 \times (3x + 2)$

$6x = 3x + 2$

Subtract $3x$ from both sides: $6x - 3x = 2$

$3x = 2$

Divide by $3$: $x = \frac{2}{3}$

This shows that the given equality $\frac{3x}{3x + 2} = \frac{1}{2}$ is not an identity; it is only true for the specific value $x = \frac{2}{3}$. It is not true for all values of $x$.

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Therefore, the given equation is incorrect as a general identity.

The common error leading to this incorrect statement is attempting to cancel the term $3x$ from the numerator and the first term of the denominator, which is invalid because $3x$ is not a factor of the entire denominator $(3x + 2)$. The incorrect cancellation looks like $\frac{\cancel{3x}}{\cancel{3x} + 2} \neq \frac{1}{2}$.

The fraction $\frac{3x}{3x + 2}$ cannot be simplified further in general.

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The given equation is $\frac{3}{4x + 3} = \frac{1}{4x}$.

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Let's analyze the Left Hand Side (LHS): LHS $= \frac{3}{4x + 3}$.

The numerator is a constant $3$. The denominator is a sum of two terms, $4x$ and $3$.

There are no common factors between the numerator ($3$) and the denominator ($4x + 3$). Note that $3$ is a term in the denominator, not a factor of the entire expression $(4x + 3)$ (unless $4x=0$, i.e., $x=0$, which would make the denominator $3$, giving $\frac{3}{3}=1$ on the LHS, and the RHS undefined).

Therefore, the fraction $\frac{3}{4x + 3}$ cannot be simplified by cancelling any terms.

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The Right Hand Side (RHS) is $\frac{1}{4x}$.

Assuming $x \neq 0$.

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To check if the equation is true for all values of $x$ (where denominators are non-zero), we can cross-multiply: $3 \times (4x) = 1 \times (4x + 3)$

$12x = 4x + 3$

Subtract $4x$ from both sides: $12x - 4x = 3$

$8x = 3$

Divide by $8$: $x = \frac{3}{8}$

This shows that the given equality $\frac{3}{4x + 3} = \frac{1}{4x}$ is not an identity; it is only true for the specific value $x = \frac{3}{8}$. It is not true for all values of $x$ (where the denominators are non-zero).

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Therefore, the given equation is incorrect as a general identity.

The common error leading to this incorrect statement is attempting to cancel or remove terms from the denominator when they are part of a sum or difference, instead of being factors of the entire denominator. One cannot cancel the constant $3$ or the term $4x$ independently in the denominator $4x+3$. The incorrect logic might be trying to somehow relate $\frac{3}{4x + 3}$ to $\frac{1}{4x}$ by cancelling or manipulating parts of the denominator.

The fraction $\frac{3}{4x + 3}$ cannot be simplified into the form $\frac{1}{4x}$.

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The given equation is $\frac{4x + 5}{4x} = 5$.

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Let's evaluate the Left Hand Side (LHS): LHS $= \frac{4x + 5}{4x}$.

When a numerator with multiple terms is divided by a single term denominator, the denominator must divide each term in the numerator.

So, we can rewrite the fraction as a sum of two fractions: LHS $= \frac{4x}{4x} + \frac{5}{4x}$

Assuming $x \neq 0$, we can simplify the first term: $\frac{4x}{4x} = \frac{\cancel{4x}}{\cancel{4x}} = 1$.

The second term $\frac{5}{4x}$ cannot be simplified further.

So, the correct simplification of the LHS is: LHS $= 1 + \frac{5}{4x}$

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The Right Hand Side (RHS) is $5$.

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Comparing the LHS and RHS: $1 + \frac{5}{4x}$ and $5$

These expressions are not equal in general. The equation $1 + \frac{5}{4x} = 5$ implies $\frac{5}{4x} = 4$, or $5 = 4 \times 4x$, which is $5 = 16x$, or $x = \frac{5}{16}$. It is only true for the specific value $x = \frac{5}{16}$ and not for all values of $x$ (where $x \neq 0$).

Therefore, the given equation is incorrect as a general identity.

The common error leading to this incorrect statement is attempting to cancel the term $4x$ from the first term of the numerator and the denominator when there is another term ($+5$) being added in the numerator. The incorrect cancellation looks like $\frac{\cancel{4x} + 5}{\cancel{4x}} \neq 1 + 5 = 6$ or simply assuming $\frac{4x+5}{4x} = \frac{5}{1} = 5$ by somehow getting rid of the $4x$ term entirely, which is invalid. The denominator must divide every term in the numerator.

The correct simplification is $1 + \frac{5}{4x}$.

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The given equation is $\frac{7x + 5}{5} = 7x$.

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Let's evaluate the Left Hand Side (LHS): LHS $= \frac{7x + 5}{5}$.

When a numerator with multiple terms is divided by a single term denominator, the denominator must divide each term in the numerator.

So, we can rewrite the fraction as a sum of two fractions: LHS $= \frac{7x}{5} + \frac{5}{5}$

Simplify the second term: $\frac{5}{5} = 1$.

So, the correct simplification of the LHS is: LHS $= \frac{7x}{5} + 1$

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The Right Hand Side (RHS) is $7x$.

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Comparing the LHS and RHS: $\frac{7x}{5} + 1$ and $7x$

These expressions are not equal in general. For example, if $x=1$: LHS $= \frac{7(1)}{5} + 1 = \frac{7}{5} + 1 = \frac{7+5}{5} = \frac{12}{5}$. RHS $= 7(1) = 7$. Since $\frac{12}{5} \neq 7$, the equation is not true for $x=1$.

The equation $\frac{7x}{5} + 1 = 7x$ implies $1 = 7x - \frac{7x}{5} = \frac{35x - 7x}{5} = \frac{28x}{5}$. This gives $5 = 28x$, so $x = \frac{5}{28}$. The equation is only true for this specific value of $x$.

Therefore, the given equation is incorrect as a general identity.

The error is in incorrectly simplifying the fraction $\frac{7x + 5}{5}$. It seems to have incorrectly cancelled the denominator $5$ with only the constant term $5$ in the numerator, ignoring the $7x$ term. An invalid cancellation like $\frac{7x + \cancel{5}}{\cancel{5}} \neq 7x$ was performed. The denominator must divide both terms in the numerator.

The correct simplification is $\frac{7x}{5} + 1$.

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