# NCERT Solutions for Class 8 Maths Chapter 12 Factorisation(Updated Pattern)

- When we factorise an expression, we write it as a product of factors. These factors may be

numbers, algebraic variables or algebraic expressions. - An irreducible factor is a factor which cannot be expressed further as a product of factors.
- A systematic way of factorising an expression is the common factor method. It consists of

three steps: (i) Write each term of the expression as a product of irreducible factors (ii) Look

for and separate the common factors and (iii) Combine the remaining factors in each term in

accordance with the distributive law. - Sometimes, all the terms in a given expression do not have a common factor; but the terms can

be grouped in such a way that all the terms in each group have a common factor. When we do

this, there emerges a common factor across all the groups leading to the required factorisation

of the expression. This is the method of regrouping. - In factorisation by regrouping, we should remember that any regrouping (i.e., rearrangement)

of the terms in the given expression may not lead to factorisation. We must observe the

expression and come out with the desired regrouping by trial and error - In expressions which have factors of the type (x + a) (x + b), remember the numerical term gives ab. Its

factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of x. - We know that in the case of numbers, division is the inverse of multiplication. This idea applies

also to the division of algebraic expressions. - In the case of the division of a polynomial by a monomial, we may carry out the division either by

dividing each term of the polynomial by the monomial or by the common factor method. - In the case of the division of a polynomial by a polynomial, we cannot proceed by dividing each term

in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials

and cancel their common factors. - In the case of divisions of algebraic expressions that we studied in this chapter, we have

Dividend = Divisor × Quotient.

In general, however, the relation is

Dividend = Divisor × Quotient + Remainder

Thus, we have considered in the present chapter only those divisions in which the remainder

is zero.

**NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportional (Updated Pattern)**

**——————————-Exercise 12.1——————————————**

**1. Find the common factors of the given terms.**

**(i) 12x, 36**

**(ii) 2y, 22xy**

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

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

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

**(vi) 16 x ^{3}, – 4x^{2} , 32 x**

**(vii) 10 pq, 20qr, 30 rp**

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

**Solution:**

(i) Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

Common factors of 12x and 36 are 2, 2, 3

and , 2×2×3 = 12

(ii) Factors of 2y and 22xy

2y = 2×y

22xy = 2×11×x×y

Common factors of 2y and 22xy are 2, y

and ,2×y = 2y

(iii) Factors of 14pq and 28p^{2}q^{2}

14pq = 2x7xpxq

28p^{2}q^{2} = 2x2x7xpxpxqxq

Common factors of 14 pq and 28 p^{2}q^{2} are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x^{2}and 4

2x = 2×x

3x^{2}= 3×x×x

4 = 2×2

Common factors of 2x, 3x^{2 }and 4 is 1.

(v) Factors of 6abc, 24ab^{2} and 12a^{2}b

6abc = 2×3×a×b×c

24ab^{2} = 2×2×2×3×a×b×b

12 a^{2 }b = 2×2×3×a×a×b

Common factors of 6 abc, 24ab^{2} and 12a^{2}b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) Factors of 16x^{3 }, -4x^{2}and 32x

16 x^{3 }= 2×2×2×2×x×x×x

– 4x^{2} = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x^{3 }, – 4x^{2 }and 32x are 2,2, x

and, 2×2×x = 4x

(vii) Factors of 10 pq, 20qr and 30rp

10 pq = 2×5×p×q

20qr = 2×2×5×q×r

30rp= 2×3×5×r×p

Common factors of 10 pq, 20qr and 30rp are 2, 5

and, 2×5 = 10

(viii) Factors of 3x^{2}y^{3} , 10x^{3}y^{2} and 6x^{2}y^{2}z

3x^{2}y^{3} = 3×x×x×y×y×y

10x^{3 }y^{2} = 2×5×x×x×x×y×y

6x^{2}y^{2}z = 3×2×x×x×y×y×z

Common factors of 3x^{2}y^{3}, 10x^{3}y^{2} and 6x^{2}y^{2}z are x^{2}, y^{2}

and, x^{2}×y^{2} = x^{2}y^{2}

**2.Factorise the following expressions.**

**(i) 7x–42**

**(ii) 6p–12q**

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

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

**(v) 20l ^{2}m+30alm**

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

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

**(viii) -4a ^{2}+4ab–4 ca**

**(ix) x ^{2}yz+xy^{2}z +xyz^{2}**

**(x) ax ^{2}y+bxy^{2}+cxyz**

**Solution:**

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

10a^{2 }= 2×5×a×a

– 15b^{2 }= -1×3×5×b×b

20c^{2} = 2×2×5×c×c

Common factor of 10 a^{2} , 15b^{2} and 20c^{2} is 5

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

(viii) – 4a^{2}+4ab-4ca

– 4a^{2} = -1×2×2×a×a

4ab = 2×2×a×b

– 4ca = -1×2×2×c×a

Common factor of – 4a^{2} , 4ab , – 4ca are 2, 2, a i.e. 4a

So,

– 4a^{2}+4 ab-4 ca = 4a(-a+b-c)

(ix) x^{2}yz+xy^{2}z+xyz^{2}

x^{2}yz = x×x×y×z

xy^{2}z = x×y×y×z

xyz^{2} = x×y×z×z

Common factor of x^{2}yz , xy^{2}z and xyz^{2} are x, y, z i.e. xyz

Now, x^{2}yz+xy^{2}z+xyz^{2 }= xyz(x+y+z)

(x) ax^{2}y+bxy^{2}+cxyz

ax^{2}y = a×x×x×y

bxy^{2 }= b×x×y×y

cxyz = c×x×y×z

Common factors of a x^{2}y ,bxy^{2 }and cxyz are xy

Now, ax^{2}y+bxy^{2}+cxyz = xy(ax+by+cz)

**3. Factorise.**

**(i) x ^{2}+xy+8x+8y**

**(ii) 15xy–6x+5y–2**

**(iii) ax+bx–ay–by**

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

**(v) z–7+7xy–xyz**

**Solution**:

**——————————-Exercise 12.2——————————————**

**1. Factorise the following expressions.**

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

**(ii) p ^{2}–10p+25**

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

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

**(v) 4x ^{2}–8x+4**

**(vi) 121b ^{2}–88bc+16c^{2}**

**(vii) (l+m) ^{2}–4lm (Hint: Expand (l+m)^{2} first)**

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

**Solution:**

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

= a^{2}+2×4×a+4^{2}

= (a+4)^{2}

Using the identity (x+y)^{2} = x^{2}+2xy+y^{2}

(ii) p^{2}–10p+25

= p^{2}-2×5×p+5^{2}

= (p-5)^{2}

Using the identity (x-y)^{2} = x^{2}-2xy+y^{2}

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

= (5m)^{2}+2×5m×3+3^{2}

= (5m+3)^{2}

Using the identity (x+y)^{2} = x^{2}+2xy+y^{2}

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

=(7y)^{2}+2×7y×6z+(6z)^{2}

= (7y+6z)^{2}

Using the identity (x+y)^{2} = x^{2}+2xy+y^{2}

(v) 4x^{2}–8x+4

= (2x)^{2}-2×4x+2^{2}

= (2x-2)^{2}

Using the identity (x-y)^{2} = x^{2}-2xy+y^{2}

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

= (11b)^{2}-2×11b×4c+(4c)^{2}

= (11b-4c)^{2}

Using the identity (x-y)^{2} = x^{2}-2xy+y^{2}

(vii) (l+m)^{2}-4lm (Hint: Expand (l+m)^{2} first)

Expand (l+m)^{2 }using the identity (x+y)^{2} = x^{2}+2xy+y^{2}

(l+m)^{2}-4lm = l^{2}+m^{2}+2lm-4lm

= l^{2}+m^{2}-2lm

= (l-m)^{2}

Using the identity (x-y)^{2} = x^{2}-2xy+y^{2}

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

= (a^{2})^{2}+2×a^{2×}b^{2}+(b^{2})^{2}

= (a^{2}+b^{2})^{2}

Using the identity (x+y)^{2} = x^{2}+2xy+y^{2}

**GLOBAL FOOTBALL FEVER: UNVEILING THE TOP 10 SOCCER LEAGUES AROUND THE WORLD**

**2. Factorise.**

**(i) 4p ^{2}–9q^{2}**

**(ii) 63a ^{2}–112b^{2}**

**(iii) 49x ^{2}–36**

**(iv) 16x ^{5}–144x^{3} differ**

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

**(vi) 9x ^{2}y^{2}–16**

**(vii) (x ^{2}–2xy+y^{2})–z^{2}**

**(viii) 25a ^{2}–4b^{2}+28bc–49c^{2}**

**Solution:**

(i) 4p^{2}–9q^{2}

= (2p)^{2}-(3q)^{2}

= (2p-3q)(2p+3q)

Using the identity x^{2}-y^{2} = (x+y)(x-y)

(ii) 63a^{2}–112b^{2}

= 7(9a^{2} –16b^{2})

= 7((3a)^{2}–(4b)^{2})

= 7(3a+4b)(3a-4b)

Using the identity x^{2}-y^{2} = (x+y)(x-y)

(iii) 49x^{2}–36

= (7x)^{2} -6^{2}

= (7x+6)(7x–6)

Using the identity x^{2}-y^{2} = (x+y)(x-y)

(iv) 16x^{5}–144x^{3}

= 16x^{3}(x^{2}–9)

= 16x^{3}(x^{2}–9)

= 16x^{3}(x–3)(x+3)

Using the identity x^{2}-y^{2} = (x+y)(x-y)

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

= {(l+m)-(l–m)}{(l +m)+(l–m)}

Using the identity x^{2}-y^{2} = (x+y)(x-y)

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

= 4 ml

(vi) 9x^{2}y^{2}–16

= (3xy)^{2}-4^{2}

= (3xy–4)(3xy+4)

Using the identity x^{2}-y^{2} = (x+y)(x-y)

(vii) (x^{2}–2xy+y^{2})–z^{2}

= (x–y)^{2}–z^{2}

Using the identity (x-y)^{2} = x^{2}-2xy+y^{2}

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using the identity x^{2}-y^{2} = (x+y)(x-y)

(viii) 25a^{2}–4b^{2}+28bc–49c^{2}

= 25a^{2}–(4b^{2}-28bc+49c^{2} )

= (5a)^{2}-{(2b)^{2}-2(2b)(7c)+(7c)^{2}}

= (5a)^{2}-(2b-7c)^{2}

Using the identity x^{2}-y^{2} = (x+y)(x-y) , we have

= (5a+2b-7c)(5a-2b+7c)

**3. Factorise the expressions.**

**(i) ax ^{2}+bx**

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

**(iii) 2x ^{3}+2xy^{2}+2xz^{2}**

**(iv) am ^{2}+bm^{2}+bn^{2}+an^{2}**

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

**(vi) y(y+z)+9(y+z)**

**(vii) 5y ^{2}–20y–8z+2yz**

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

**(ix)6xy–4y+6–9x**

**Solution:**

(i) ax^{2}+bx = x(ax+b)

(ii) 7p^{2}+21q^{2} = 7(p^{2}+3q^{2})

(iii) 2x^{3}+2xy^{2}+2xz^{2} = 2x(x^{2}+y^{2}+z^{2})

(iv) am^{2}+bm^{2}+bn^{2}+an^{2 }= m^{2}(a+b)+n^{2}(a+b) = (a+b)(m^{2}+n^{2})

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

(vi) y(y+z)+9(y+z) = (y+9)(y+z)

(vii) 5y^{2}–20y–8z+2yz = 5y(y–4)+2z(y–4) = (y–4)(5y+2z)

(viii) 10ab+4a+5b+2 = 5b(2a+1)+2(2a+1) = (2a+1)(5b+2)

(ix) 6xy–4y+6–9x = 6xy–9x–4y+6 = 3x(2y–3)–2(2y–3) = (2y–3)(3x–2)

**4.Factorise.**

**(i) a ^{4}–b^{4}**

**(ii) p ^{4}–81**

**(iii) x ^{4}–(y+z)^{ 4}**

**(iv) x ^{4}–(x–z)^{ 4}**

**(v) a ^{4}–2a^{2}b^{2}+b^{4}**

**Solution:**

(i) a^{4}–b^{4}

= (a^{2})^{2}-(b^{2})^{2}

= (a^{2}-b^{2}) (a^{2}+b^{2})

= (a – b)(a + b)(a^{2}+b^{2})

(ii) p^{4}–81

= (p^{2})^{2}-(9)^{2}

= (p^{2}-9)(p^{2}+9)

= (p^{2}-3^{2})(p^{2}+9)

=(p-3)(p+3)(p^{2}+9)

(iii) x^{4}–(y+z)^{ 4} = (x^{2})^{2}-[(y+z)^{2}]^{2}

= {x^{2}-(y+z)^{2}}{ x^{2}+(y+z)^{2}}

= {(x –(y+z)(x+(y+z)}{x^{2}+(y+z)^{2}}

= (x–y–z)(x+y+z) {x^{2}+(y+z)^{2}}

(iv) x^{4}–(x–z)^{ 4} = (x^{2})^{2}-{(x-z)^{2}}^{2}

= {x^{2}-(x-z)^{2}}{x^{2}+(x-z)^{2}}

= { x-(x-z)}{x+(x-z)} {x^{2}+(x-z)^{2}}

= z(2x-z)( x^{2}+x^{2}-2xz+z^{2})

= z(2x-z)( 2x^{2}-2xz+z^{2})

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

= (a^{2}-b^{2})^{2}

= ((a–b)(a+b))^{2}

= (a – b)^{2} (a + b)^{2}

**5. Factorise the following expressions.**

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

**(ii) q ^{2}–10q+21**

**(iii) p ^{2}+6p–16**

**Solution:**

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

We observed that 8 = 4×2 and 4+2 = 6

p^{2}+6p+8 can be written as p^{2}+2p+4p+8

Taking Common terms, we get

p^{2}+6p+8 = p^{2}+2p+4p+8 = p(p+2)+4(p+2)

Again, p+2 is common in both the terms.

= (p+2)(p+4)

This implies that p^{2}+6p+8 = (p+2)(p+4)

(ii) q^{2}–10q+21

We observed that 21 = -7×-3 and -7+(-3) = -10

q^{2}–10q+21 = q^{2}–3q-7q+21

= q(q–3)–7(q–3)

= (q–7)(q–3)

This implies that q^{2}–10q+21 = (q–7)(q–3)

(iii) p^{2}+6p–16

We observed that -16 = -2×8 and 8+(-2) = 6

p^{2}+6p–16 = p^{2}–2p+8p–16

= p(p–2)+8(p–2)

= (p+8)(p–2)

So, p^{2}+6p–16 = (p+8)(p–2)

**——————————-Exercise 12.3——————————————**

**1. Carry out the following divisions.**

**(i) 28x ^{4} ÷ 56x**

**(ii) –36y ^{3} ÷ 9y^{2}**

**(iii) 66pq ^{2}r^{3} ÷ 11qr^{2}**

**(iv) 34x ^{3}y^{3}z^{3} ÷ 51xy^{2}z^{3}**

**(v) 12a ^{8}b^{8} ÷ (– 6a^{6}b^{4})**

**Solution:**

(i)28x^{4} = 2×2×7×x×x×x×x

56x = 2×2×2×7×x

**2. Divide the given polynomial by the given monomial.**

**(i)(5x ^{2}–6x) ÷ 3x**

**(ii)(3y ^{8}–4y^{6}+5y^{4}) ÷ y^{4}**

**(iii) 8(x ^{3}y^{2}z^{2}+x^{2}y^{3}z^{2}+x^{2}y^{2}z^{3})÷ 4x^{2 }y^{2 }z^{2}**

**(iv)(x ^{3}+2x^{2}+3x) ÷2x**

**(v) (p ^{3}q^{6}–p^{6}q^{3}) ÷ p^{3}q^{3}**

**Solution:**

**3. Work out the following divisions.**

**(i) (10x–25) ÷ 5**

**(ii) (10x–25) ÷ (2x–5)**

**(iii) 10y(6y+21) ÷ 5(2y+7)**

**(iv) 9x ^{2}y^{2}(3z–24) ÷ 27xy(z–8)**

**(v) 96abc(3a–12)(5b–30) ÷ 144(a–4)(b–6)**

**Solution:**

(i) (10x–25) ÷ 5 = 5(2x-5)/5 = 2x-5

(ii) (10x–25) ÷ (2x–5) = 5(2x-5)/( 2x-5) = 5

(iii) 10y(6y+21) ÷ 5(2y+7) = 10y×3(2y+7)/5(2y+7) = 6y

(iv) 9x^{2}y^{2}(3z–24) ÷ 27xy(z–8) = 9x^{2}y^{2}×3(z-8)/27xy(z-8) = xy

**4. Divide as directed.**

**(i) 5(2x+1)(3x+5)÷ (2x+1)**

**(ii) 26xy(x+5)(y–4)÷13x(y–4)**

**(iii) 52pqr(p+q)(q+r)(r+p) ÷ 104pq(q+r)(r+p)**

**(iv) 20(y+4) (y ^{2}+5y+3) ÷ 5(y+4)**

**(v) x(x+1) (x+2)(x+3) ÷ x(x+1)**

**Solution:**

**5. Factorise the expressions and divide them as directed.**

**(i) (y ^{2}+7y+10)÷(y+5)**

**(ii) (m ^{2}–14m–32)÷(m+2)**

**(iii) (5p ^{2}–25p+20)÷(p–1)**

**(iv) 4yz(z ^{2}+6z–16)÷2y(z+8)**

**(v) 5pq(p ^{2}–q^{2})÷2p(p+q)**

**(vi) 12xy(9x ^{2}–16y^{2})÷4xy(3x+4y)**

**(vii) 39y ^{3}(50y^{2}–98) ÷ 26y^{2}(5y+7)**

**Solution:**

**(i) (y ^{2}+7y+10)÷(y+5)**

First, solve the equation (y^{2}+7y+10)

(y^{2}+7y+10) = y^{2}+2y+5y+10 = y(y+2)+5(y+2) = (y+2)(y+5)

Now, (y^{2}+7y+10)÷(y+5) = (y+2)(y+5)/(y+5) = y+2

**(ii) (m ^{2}–14m–32)÷ (m+2)**

Solve for m^{2}–14m–32, we have

m^{2}–14m–32 = m^{2}+2m-16m–32 = m(m+2)–16(m+2) = (m–16)(m+2)

Now, (m^{2}–14m–32)÷(m+2) = (m–16)(m+2)/(m+2) = m-16

**(iii) (5p ^{2}–25p+20)÷(p–1)**

Step 1: Take 5 common from the equation, 5p^{2}–25p+20, we get

5p^{2}–25p+20 = 5(p^{2}–5p+4)

Step 2: Factorise p^{2}–5p+4

p^{2}–5p+4 = p^{2}–p-4p+4 = (p–1)(p–4)

Step 3: Solve original equation

(5p^{2}–25p+20)÷(p–1) = 5(p–1)(p–4)/(p-1) = 5(p–4)

**(iv) 4yz(z ^{2} + 6z–16)÷ 2y(z+8)**

Factorising z^{2}+6z–16,

z^{2}+6z–16 = z^{2}-2z+8z–16 = (z–2)(z+8)

Now, 4yz(z^{2}+6z–16) ÷ 2y(z+8) = 4yz(z–2)(z+8)/2y(z+8) = 2z(z-2)

**(v) 5pq(p ^{2}–q^{2}) ÷ 2p(p+q)**

p^{2}–q^{2} can be written as (p–q)(p+q) using the identity.

5pq(p^{2}–q^{2}) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5q(p–q)/2

**(vi) 12xy(9x ^{2}–16y^{2}) ÷ 4xy(3x+4y)**

Factorising 9x^{2}–16y^{2} , we have

9x^{2}–16y^{2} = (3x)^{2}–(4y)^{2} = (3x+4y)(3x-4y) using the identity p^{2}–q^{2} = (p–q)(p+q)

Now, 12xy(9x^{2}–16y^{2}) ÷ 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)

**(vii) 39y ^{3}(50y^{2}–98) ÷ 26y^{2}(5y+7)**

st solve for 50y

^{2}–98, we have

50y^{2}–98 = 2(25y^{2}–49) = 2((5y)^{2}–7^{2}) = 2(5y–7)(5y+7)

Now, 39y^{3}(50y^{2}–98) ÷ 26y^{2}(5y+7) =