# Factorisation - Class 8 - Mathematics

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## Examples - Factorisation | NCERT | Mathematics | Class 8

Factorise $12 a^{2} b+15 a b^{2}$

To factorize the expression $12a^{2}b + 15ab^{2}$, follow these steps:

**Find the Greatest Common Factor (GCF):**Identify the GCF of the coefficients (numeric factors), which in this case is 3.

Also, find the GCF of the variables, which is $a$ (with the lowest exponent of 1) and $b$ (also with the lowest exponent of 1).

**Rewrite the Expression Using the GCF:**Factor out the GCF from each term.

Putting it all together:

**Step 1:**The GCF of $12, 15$ is $3$, and the least powers of $a$ and $b$ present in both terms are $a^1$ and $b^1$, respectively. So, the GCF is $3ab$.**Step 2:**Divide each term by $3ab$:$12a^{2}b$ becomes $4a$, because $12a^{2}b / 3ab = 4a$.

$15ab^{2}$ becomes $5b$, because $15ab^{2} / 3ab = 5b$.

**Final Factored Form:**$$3ab(4a + 5b)$$

Factorise $10 x^{2}-18 x^{3}+14 x^{4}$

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Sign up nowFactorise $6 x y-4 y+6-9 x$

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Sign up nowFactorise $x^{2}+8 x+16$

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Sign up nowFactorise $4 y^{2}-12 y+9$

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Factorise $49 p^{2}-36$

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Factorise $a^{2}-2 a b+b^{2}-c^{2}$

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Factorise $m^{4}-256$

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Factorise $x^{2}+5 x+6$

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Find the factors of $y^{2}-7 y+12$

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Obtain the factors of $z^{2}-4 z-12$

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Find the factors of $3 m^{2}+9 m+6$

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Do the following divisions.

(i) $-20 x^{4} \div 10 x^{2}$

(ii) $7 x^{2} y^{2} z^{2} \div 14 x y z$

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Divide $44\left(x^{4}-5 x^{3}-24 x^{2}\right)$ by $11 x(x-8)$

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## Exercise 12.1 - Factorisation | NCERT | Mathematics | Class 8

Find the common factors of the given terms

(i) $12 x, 36$

(ii) $2 y, 22 x y$

(iii) $14 p q, 28 p^{2} q^{2}$

(iv) $2 x, 3 x^{2}, 4$

(v) $6 a b c, 24 a b^{2}, 12 a^{2} b$

(vi) $16 x^{3},-4 x^{2}, 32 x$

(vii) $10 p q, 20 q r, 30 r p$

(viii) $3 x^{2} y^{3}, 10 x^{3} y^{2}, 6 x^{2} y^{2} z$

To find the common factors of the given terms, let's examine each item step by step.

(i) **Terms:** $12x, 36$

Prime factorization $12 = 2^2 \cdot 3$, and $36 = 2^2 \cdot 3^2$.

**Common factors:**$2^2$ and $3$.

(ii) **Terms:** $2y, 22xy$

Prime factorization $2 = 2$, and $22 = 2 \cdot 11$.

**Common factors:**$2$ and incorporating variable, $2y$ if considering y in both.

(iii) **Terms:** $14pq, 28p^{2}q^{2}$

Prime factorization $14 = 2 \cdot 7$, and $28 = 2^2 \cdot 7$.

**Common factors:**$2$, $7$, and incorporating variables, $2pq$.

(iv) **Terms:** $2x, 3x^{2}, 4$

Prime factorization $2 = 2$, $3 = 3$, and $4 = 2^2$.

**Common factors:**The constant factor is $1$, and for variables, $x$ is common.

(v) **Terms:** $6abc, 24ab^{2}, 12a^{2}b$

Prime factorization $6 = 2 \cdot 3$, $24 = 2^3 \cdot 3$, and $12 = 2^2 \cdot 3$.

**Common factors:**$2$ and $3$, for variables $a$ and $b$ are common, so $6ab$.

(vi) **Terms:** $16x^{3}, -4x^{2}, 32x$

Prime factorization $16 = 2^4$, $4 = 2^2$, and $32 = 2^5$.

**Common factors:**Considering the sign, just the magnitude, $2^2$ is common, and for variables, $x$. Thus, $4x$.

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

Prime factorization $10 = 2 \cdot 5$, $20 = 2^2 \cdot 5$, and $30 = 2 \cdot 3 \cdot 5$.

**Common factors:**No constant number is common across all three, and there's no single variable present in all; hence, no common factors.

(viii) **Terms:** $3x^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}z$

Prime factorization $3 = 3$, $10 = 2 \cdot 5$, and $6 = 2 \cdot 3$.

**Common factors:**For numbers, no common factor across all three. For variables, $x^2$ and $y^2$ are common.

These steps bring us to the conclusion about the common factors for each set of terms.

Factorise the following expressions

(i) $7 x-42$

(ii) $6 p-12 q$

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

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

(v) $20 l^{2} m+30 a l m$

(vi) $5 x^{2} y-15 x y^{2}$

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

(viii) $-4 a^{2}+4 a b-4 c a$

(ix) $x^{2} y z+x y^{2} z+x y z^{2}$

(x) $a x^{2} y+b x y^{2}+c x y z$

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Factorise

(i) $x^{2}+x y+8 x+8 y$

(ii) $15 x y-6 x+5 y-2$

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

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

(v) $z-7+7 x y-x y z$

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## Exercise 12.2 - Factorisation | NCERT | Mathematics | Class 8

Factorise the following expressions

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

(ii) $p^{2}-10 p+25$

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

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

(v) $4 x^{2}-8 x+4$

(vi) $121 b^{2}-88 b c+16 c^{2}$

(vii) $(l+m)^{2}-4 l m$

(Hint: Expand $(l+m)^{2}$ first)

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

Let's go through the factorization of each expression step by step.

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

This is a perfect square trinomial. The pattern is:

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

**Step 1:**Identify $a$ and $b$.**Step 2:**Apply the pattern.

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

### (ii) $p^2 - 10p + 25$

This is another perfect square trinomial.

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

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

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

This is also a perfect square trinomial.

$$ (5m)^2 + 2(5m)(3) + 3^2 = (5m + 3)^2 $$

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

This can be factored as a perfect square trinomial.

$$ (7y)^2 + 2(7y)(6z) + (6z)^2 = (7y + 6z)^2 $$

### (v) $4x^2 - 8x + 4$

Another perfect square trinomial.

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

### (vi) $121b^2 - 88bc + 16c^2$

This also forms a perfect square trinomial.

$$ (11b)^2 - 2(11b)(4c) + (4c)^2 = (11b - 4c)^2 $$

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

This leverages the difference of two squares:

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

**Step 1:**Expand $(l + m)^2$.**Step 2:**Notice $(l + m)^2 = l^2 + 2lm + m^2$.

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

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

This can be identified as a perfect square as well:

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

Thus, each expression is factored by recognizing them as perfect squares or by applying algebraic identities.

Factorise

(i) $4 p^{2}-9 q^{2}$

(ii) $63 a^{2}-112 b^{2}$

(iii) $49 x^{2}-36$

(iv) $16 x^{5}-144 x^{3}$

(v) $(l+m)^{2}-(l-m)^{2}$

(vi) $9 x^{2} y^{2}-16$

(vii) $\left(x^{2}-2 x y+y^{2}\right)-z^{2}$

(viii) $25 a^{2}-4 b^{2}+28 b c-49 c^{2}$

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Factorise the expressions

(i) $a x^{2}+b x$

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

(iii) $2 x^{3}+2 x y^{2}+2 x z^{2}$

(iv) $a m^{2}+b m^{2}+b n^{2}+a n^{2}$

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

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

(vii) $5 y^{2}-20 y-8 z+2 y z$

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

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

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Factorise

(i) $a^{4}-b^{4}$

(iv) $x^{4}-(x-z)^{4}$

(ii) $p^{4}-81$

(v) $a^{4}-2 a^{2} b^{2}+b^{4}$

(iii) $x^{4}-(y+z)^{4}$

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Factorise the following expressions

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

(ii) $q^{2}-10 q+21$

(iii) $p^{2}+6 p-16$

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## Exercise 12.3 - Factorisation | NCERT | Mathematics | Class 8

Carry out the following divisions

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

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

(iii) $66 p q^{2} r^{3} \div 11 q r^{2}$

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

(v) $12 a^{8} b^{8} \div\left(-6 a^{6} b^{4}\right)$

To carry out these divisions with steps, we divide both the coefficients and subtract the exponents for like bases:

### (i) $28 x^{4} \div 56 x$

Divide the coefficients: $28 \div 56 = 1/2$

Subtract the exponents of like bases: $x^{4} \div x^{1} = x^{4-1} = x^{3}$

So, the result is $\frac{1}{2} x^{3}$.

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

Divide the coefficients: $-36 \div 9 = -4$

Subtract the exponents of like bases: $y^{3} \div y^{2} = y^{3-2} = y^{1} = y$

So, the result is $-4y$.

### (iii) $66 p q^{2} r^{3} \div 11 q r^{2}$

Divide the coefficients: $66 \div 11 = 6$

For $p$, since there is no matching base in the divisor, it remains as $p$

Subtract the exponents of $q$: $q^{2} \div q = q^{2-1} = q^{1} = q$

Subtract the exponents of $r$: $r^{3} \div r^{2} = r^{3-2} = r^{1} = r$

So, the result is $6 p q r$.

### (iv) $34 x^{3} y^{3} z^{3} \div 51 x y^{2} z^{3}$

Divide the coefficients: $34 \div 51 = 2/3$

Subtract the exponents of $x$: $x^{3} \div x = x^{3-1} = x^{2}$

Subtract the exponents of $y$: $y^{3} \div y^{2} = y^{3-2} = y^{1} = y$

For $z$, since the exponents are equal ($z^{3} \div z^{3}$), they cancel out to $1$

So, the result is $\frac{2}{3} x^{2} y$.

### (v) $12 a^{8} b^{8} \div\left(-6 a^{6} b^{4}\right)$

Divide the coefficients: $12 \div -6 = -2$

Subtract the exponents of $a$: $a^{8} \div a^{6} = a^{8-6} = a^{2}$

Subtract the exponents of $b$: $b^{8} \div b^{4} = b^{8-4} = b^{4}$

So, the result is $-2 a^{2} b^{4}$.

Divide the given polynomial by the given monomial

(i) $\left(5 x^{2}-6 x\right) \div 3 x$

(ii) $\left(3 y^{8}-4 y^{6}+5 y^{4}\right) \div y^{4}$

(iii) $8\left(x^{3} y^{2} z^{2}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3}\right) \div 4 x^{2} y^{2} z^{2}$

(iv) $\left(x^{3}+2 x^{2}+3 x\right) \div 2 x$

(v) $\left(p^{3} q^{6}-p^{6} q^{3}\right) \div p^{3} q^{3}$

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Work out the following divisions

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

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

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

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

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

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Divide as directed

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

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

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

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

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

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Factorise the expressions and divide them as directed

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

(ii) $\left(m^{2}-14 m-32\right) \div(m+2)$

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

(v) $5 p q\left(p^{2}-q^{2}\right) \div 2 p(p+q)$

(iv) $4 y z\left(z^{2}+6 z-16\right) \div 2 y(z+8)$

(vi) $12 x y\left(9 x^{2}-16 y^{2}\right) \div 4 x y(3 x+4 y)$

(vii) $39 y^{3}\left(50 y^{2}-98\right) \div 26 y^{2}(5 y+7)$