# NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

The NCERT Solutions are aimed at mastering the concepts and acquiring comprehensive knowledge about the various types of questions asked in CBSE Class 8 Mathematics Examinations. All the questions are provided with appropriate solutions which will come in handy to revise the CBSE class 8 Mathematics syllabus. Solutions for Maths Class 8 are designed by highly knowledgeable subject experts as per the CBSE curriculum. Class 8 Maths NCERT Solutions are helpful for the students to streamline their last-minute revision. Students can download pdf and study offline.

## Download PDF of NCERT Solutions for Class 8 Maths Chapter 14 Factorisation ## Exercise 14.1 Page No: 208

1. Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14 pq, 28p2q2

(iv) 2x, 3x2, 4

(v) 6 abc, 24ab2, 12a2b

(vi) 16 x3, – 4x2 , 32 x

(vii) 10 pq, 20qr, 30 rp

(viii) 3x2y3 , 10x3y2 , 6x2y2z

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 28p2q

14pq = 2x7xpxq

28p2q = 2x2x7xpxpxq

Common factors of 14 pq and 28 p2q are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x2and 4

2x = 2×x

3x2= 3×x×x

4 = 2×2

Common factors of 2x, 3x2 and 4 is 1.

(v) Factors of 6abc, 24ab2 and 12a2b

6abc = 2×3×a×b×c

24ab2 = 2×2×2×3×a×b×b

12 a2 b = 2×2×3×a×a×b

Common factors of 6 abc, 24ab2 and 12a2b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) Factors of 16x3 , -4x2and 32x

16 x3 = 2×2×2×2×x×x×x

– 4x2 = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x3 , – 4x2 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 3x2y3 , 10x3y2 and 6x2y2z

3x2y3 = 3×x×x×y×y×y

10x3 y2 = 2×5×x×x×x×y×y

6x2y2z = 3×2×x×x×y×y×z

Common factors of 3x2y3, 10x3y2 and 6x2y2z are x2, y2

and, x2×y2 = x2y2

2.Factorise the following expressions

(i) 7x–42

(ii) 6p–12q

(iii) 7a2+ 14a

(iv) -16z+20 z3

(v) 20l2m+30alm

(vi) 5x2y-15xy2

(vii) 10a2-15b2+20c2

(viii) -4a2+4ab–4 ca

(ix) x2yz+xy2z +xyz2

(x) ax2y+bxy2+cxyz

Solution: (vii) 10a2-15b2+20c2

10a2 = 2×5×a×a

– 15b2 = -1×3×5×b×b

20c2 = 2×2×5×c×c

Common factor of 10 a2 , 15b2 and 20c2 is 5

10a2-15b2+20c2 = 5(2a2-3b2+4c2 )

(viii) – 4a2+4ab-4ca

– 4a2 = -1×2×2×a×a

4ab = 2×2×a×b

– 4ca = -1×2×2×c×a

Common factor of – 4a2 , 4ab , – 4ca are 2, 2, a i.e. 4a

So,

– 4a2+4 ab-4 ca = 4a(-a+b-c)

(ix) x2yz+xy2z+xyz2

x2yz = x×x×y×z

xy2z = x×y×y×z

xyz2 = x×y×z×z

Common factor of x2yz , xy2z and xyz2 are x, y, z i.e. xyz

Now, x2yz+xy2z+xyz2 = xyz(x+y+z)

(x) ax2y+bxy2+cxyz

ax2y = a×x×x×y

bxy2 = b×x×y×y

cxyz = c×x×y×z

Common factors of a x2y ,bxy2 and cxyz are xy

Now, ax2y+bxy2+cxyz = xy(ax+by+cz)

3. Factorise.

(i) x2+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 14.2 Page No: 223

1. Factorise the following expressions.

(i) a2+8a+16

(ii) p2–10p+25

(iii) 25m2+30m+9

(iv) 49y2+84yz+36z2

(v) 4x2–8x+4

(vi) 121b2–88bc+16c2

(vii) (l+m)2–4lm (Hint: Expand (l+m)2 first)

(viii) a4+2a2b2+b4

Solution:

(i) a2+8a+16

= a2+2×4×a+42

= (a+4)2

Using identity: (x+y)2 = x2+2xy+y2

(ii) p2–10p+25

= p2-2×5×p+52

= (p-5)2

Using identity: (x-y)2 = x2-2xy+y2

(iii) 25m2+30m+9

= (5m)2-2×5m×3+32

= (5m+3)2

Using identity: (x+y)2 = x2+2xy+y2

(iv) 49y2+84yz+36z2

=(7y)2+2×7y×6z+(6z)2

= (7y+6z)2

Using identity: (x+y)2 = x2+2xy+y2

(v) 4x2–8x+4

= (2x)2-2×4x+22

= (2x-2)2

Using identity: (x-y)2 = x2-2xy+y2

(vi) 121b2-88bc+16c2

= (11b)2-2×11b×4c+(4c)2

= (11b-4c)2

Using identity: (x-y)2 = x2-2xy+y2

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

Expand (l+m)2 using identity: (x+y)2 = x2+2xy+y2

(l+m)2-4lm = l2+m2+2lm-4lm

= l2+m2-2lm

= (l-m)2

Using identity: (x-y)2 = x2-2xy+y2

(viii) a4+2a2b2+b4

= (a2)2+2×ab2+(b2)2

= (a2+b2)2

Using identity: (x+y)2 = x2+2xy+y2

2. Factorise.

(i) 4p2–9q2

(ii) 63a2–112b2

(iii) 49x2–36

(iv) 16x5–144x3 differ

(v) (l+m)2-(l-m) 2

(vi) 9x2y2–16

(vii) (x2–2xy+y2)–z2

(viii) 25a2–4b2+28bc–49c2

Solution:

(i) 4p2–9q2

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

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

Using identity: x2-y2 = (x+y)(x-y)

(ii) 63a2–112b2

= 7(9a2 –16b2)

= 7((3a)2–(4b)2)

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

Using identity: x2-y2 = (x+y)(x-y)

(iii) 49x2–36

= (7a)2 -62

= (7a+6)(7a–6)

Using identity: x2-y2 = (x+y)(x-y)

(iv) 16x5–144x3

= 16x3(x2–9)

= 16x3(x2–9)

= 16x3(x–3)(x+3)

Using identity: x2-y2 = (x+y)(x-y)

(v) (l+m) 2-(l-m) 2

= {(l+m)-(l–m)}{(l +m)+(l–m)}

Using Identity: x2-y2 = (x+y)(x-y)

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

= 4 ml

(vi) 9x2y2–16

= (3xy)2-42

= (3xy–4)(3xy+4)

Using Identity: x2-y2 = (x+y)(x-y)

(vii) (x2–2xy+y2)–z2

= (x–y)2–z2

Using Identity: (x-y)2 = x2-2xy+y2

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using Identity: x2-y2 = (x+y)(x-y)

(viii) 25a2–4b2+28bc–49c2

= 25a2–(4b2-28bc+49c2 )

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

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

Using Identity: x2-y2 = (x+y)(x-y) , we have

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

3. Factorise the expressions.

(i) ax2+bx

(ii) 7p2+21q2

(iii) 2x3+2xy2+2xz2

(iv) am2+bm2+bn2+an2

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

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

(vii) 5y2–20y–8z+2yz

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

(ix)6xy–4y+6–9x

Solution:

(i) ax2+bx = x(ax+b)

(ii) 7p2+21q2 = 7(p2+3q2)

(iii) 2x3+2xy2+2xz2 = 2x(x2+y2+z2)

(iv) am2+bm2+bn2+an2 = m2(a+b)+n2(a+b) = (a+b)(m2+n2)

(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) 5y2–20y–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) a4–b4

(ii) p4–81

(iii) x4–(y+z) 4

(iv) x4–(x–z) 4

(v) a4–2a2b2+b4

Solution:

(i) a4–b4

= (a2)2-(b2)2

= (a2-b2) (a2+b2)

= (a – b)(a + b)(a2+b2)

(ii) p4–81

= (p2)2-(9)2

= (p2-9)(p2+9)

= (p2-32)(p2+9)

=(p-3)(p+3)(p2+9)

(iii) x4–(y+z) 4 = (x2)2-[(y+z)2]2

= {x2-(y+z)2}{ x2+(y+z)2}

= {(x –(y+z)(x+(y+z)}{x2+(y+z)2}

= (x–y–z)(x+y+z) {x2+(y+z)2}

(iv) x4–(x–z) 4 = (x2)2-{(x-z)2}2

= {x2-(x-z)2}{x2+(x-z)2}

= { x-(x-z)}{x+(x-z)} {x2+(x-z)2}

= z(2x-z)( x2+x2-2xz+z2)

= z(2x-z)( 2x2-2xz+z2)

(v) a4–2a2b2+b4 = (a2)2-2a2b2+(b2)2

= (a2-b2)2

= ((a–b)(a+b))2

5. Factorise the following expressions.

(i) p2+6p+8

(ii) q2–10q+21

(iii) p2+6p–16

Solution:

(i) p2+6p+8

We observed that, 8 = 4×2 and 4+2 = 6

p2+6p+8 can be written as p2+2p+4p+8

Taking Common terms, we get

p2+6p+8 = p2+2p+4p+8 = p(p+2)+4(p+2)

Again p+2 is common in both the terms.

= (p+2)(p+4)

This implies: p2+6p+8 = (p+2)(p+4)

(ii) q2–10q+21

Observed that, 21 = -7×-3 and -7+(-3) = -10

q2–10q+21 = q2–3q-7q+21

= q(q–3)–7(q–3)

= (q–7)(q–3)

This implies q2–10q+21 = (q–7)(q–3)

(iii) p2+6p–16

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

p2+6p–16 = p2–2p+8p–16

= p(p–2)+8(p–2)

= (p+8)(p–2)

So, p2+6p–16 = (p+8)(p–2)

## Exercise 14.3 Page No: 227

1. Carry out the following divisions.

(i) 28x4 ÷ 56x

(ii) –36y3 ÷ 9y2

(iii) 66pq2r3 ÷ 11qr2

(iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (– 6a6b4)

Solution:

(i)28x4 = 2×2×7×x×x×x×x

56x = 2×2×2×7×x 2. Divide the given polynomial by the given monomial.

(i)(5x2–6x) ÷ 3x

(ii)(3y8–4y6+5y4) ÷ y4

(iii) 8(x3y2z2+x2y3z2+x2y2z3)÷ 4x2 y2 z2

(iv)(x3+2x2+3x) ÷2x

(v) (p3q6–p6q3) ÷ p3q3

Solution: 3. Work out the following divisions.

(i) (10x–25) ÷ 5

(ii) (10x–25) ÷ (2x–5)

(iii) 10y(6y+21) ÷ 5(2y+7)

(iv) 9x2y2(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) 9x2y2(3z–24) ÷ 27xy(z–8) = 9x2y2×3(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) (y2+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) (y2+7y+10)÷(y+5)

(ii) (m2–14m–32)÷(m+2)

(iii) (5p2–25p+20)÷(p–1)

(iv) 4yz(z2+6z–16)÷2y(z+8)

(v) 5pq(p2–q2)÷2p(p+q)

(vi) 12xy(9x2–16y2)÷4xy(3x+4y)

(vii) 39y3(50y2–98) ÷ 26y2(5y+7)

Solution:

(i) (y2+7y+10)÷(y+5)

First solve for equation, (y2+7y+10)

(y2+7y+10) = y2+2y+5y+10 = y(y+2)+5(y+2) = (y+2)(y+5)

Now, (y2+7y+10)÷(y+5) = (y+2)(y+5)/(y+5) = y+2

(ii) (m2–14m–32)÷ (m+2)

Solve for m2–14m–32, we have

m2–14m–32 = m2+2m-16m–32 = m(m+2)–16(m+2) = (m–16)(m+2)

Now, (m2–14m–32)÷(m+2) = (m–16)(m+2)/(m+2) = m-16

(iii) (5p2–25p+20)÷(p–1)

Step 1: Take 5 common from the equation, 5p2–25p+20, we get

5p2–25p+20 = 5(p2–5p+4)

Step 2: Factorize p2–5p+4

p2–5p+4 = p2–p-4p+4 = (p–1)(p–4)

Step 3: Solve original equation

(5p2–25p+20)÷(p–1) = 5(p–1)(p–4)/(p-1) = 5(p–4)

(iv) 4yz(z2 + 6z–16)÷ 2y(z+8)

Factorize z2+6z–16,

z2+6z–16 = z2-2z+8z–16 = (z–2)(z+8)

Now, 4yz(z2+6z–16) ÷ 2y(z+8) = 4yz(z–2)(z+8)/2y(z+8) = 2z(z-2)

(v) 5pq(p2–q2) ÷ 2p(p+q)

p2–q2 can be written as (p–q)(p+q) using identity.

5pq(p2–q2) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5/2q(p–q)

(vi) 12xy(9x2–16y2) ÷ 4xy(3x+4y)

Factorize 9x2–16y2 , we have

9x2–16y2 = (3x)2–(4y)2 = (3x+4y)(3x-4y) using identity: p2–q2 = (p–q)(p+q)

Now, 12xy(9x2–16y2) ÷ 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)

(vii) 39y3(50y2–98) ÷ 26y2(5y+7)
st solve for 50y2–98, we have

50y2–98 = 2(25y2–49) = 2((5y)2–72) = 2(5y–7)(5y+7)

Now, 39y3(50y2–98) ÷ 26y2(5y+7) = ## Exercise 14.4 Page No: 228

1. 4(x–5) = 4x–5

Solution:

4(x- 5)= 4x – 20 ≠ 4x – 5 = RHS

The correct statement is 4(x-5) = 4x–20

2. x(3x+2) = 3x2+2

Solution:

LHS = x(3x+2) = 3x2+2x ≠ 3x2+2 = RHS

The correct solution is x(3x+2) = 3x2+2x

3. 2x+3y = 5xy

Solution:

LHS= 2x+3y ≠ R. H. S

The correct statement is 2x+3y = 2x+3 y

4. x+2x+3x = 5x

Solution:

LHS = x+2x+3x = 6x ≠ RHS

The correct statement is x+2x+3x = 6x

5. 5y+2y+y–7y = 0

Solution:

LHS = 5y+2y+y–7y = y ≠ RHS

The correct statement is 5y+2y+y–7y = y

6. 3x+2x = 5x2

Solution:

LHS = 3x+2x = 5x ≠ RHS

The correct statement is 3x+2x = 5x

7. (2x) 2+4(2x)+7 = 2x2+8x+7

Solution:

LHS = (2x) 2+4(2x)+7 = 4x2+8x+7 ≠ RHS

The correct statement is (2x) 2+4(2x)+7 = 4x2+8x+7

8. (2x) 2+5x = 4x+5x = 9x

Solution:

LHS = (2x) 2+5x = 4x2+5x ≠ 9x = RHS

The correct statement is(2x) 2+5x = 4x2+5x

9. (3x + 2) 2 = 3x2+6x+4

Solution:

LHS = (3x+2) 2 = (3x)2+22+2x2x3x = 9x2+4+12x ≠ RHS

The correct statement is (3x + 2) 2 = 9x2+4+12x

10. Substituting x = – 3 in

(a) x2 + 5x + 4 gives (– 3) 2+5(– 3)+4 = 9+2+4 = 15

(b) x2 – 5x + 4 gives (– 3) 2– 5( – 3)+4 = 9–15+4 = – 2

(c) x2 + 5x gives (– 3) 2+5(–3) = – 9–15 = – 24

Solution:

(a) Substituting x = – 3 in x2+5x+4, we have

x2+5x+4 = (– 3) 2+5(– 3)+4 = 9–15+4 = – 2. This is the correct answer.

(b) Substituting x = – 3 in x2–5x+4

x2–5x+4 = (–3) 2–5(– 3)+4 = 9+15+4 = 28. This is the correct answer

(c)Substituting x = – 3 in x2+5x

x2+5x = (– 3) 2+5(–3) = 9–15 = -6. This is the correct answer

11.(y–3)2 = y2–9

Solution:

LHS = (y–3)2 , which is similar to (a–b)2 identity, where (a–b) 2 = a2+b2-2ab.

(y – 3)2 = y2+(3) 2–2y×3 = y2+9 –6y ≠ y2 – 9 = RHS

The correct statement is (y–3)2 = y2 + 9 – 6y

12. (z+5) 2 = z2+25

Solution:

LHS = (z+5)2 , which is similar to (a +b)2 identity, where (a+b) 2 = a2+b2+2ab.

(z+5) 2 = z2+52+2×5×z = z2+25+10z ≠ z2+25 = RHS

The correct statement is (z+5) 2 = z2+25+10z

13. (2a+3b)(a–b) = 2a2–3b2

Solution:

LHS = (2a+3b)(a–b) = 2a(a–b)+3b(a–b)

= 2a2–2ab+3ab–3b2

= 2a2+ab–3b2

≠ 2a2–3b2 = RHS

The correct statement is (2a +3b)(a –b) = 2a2+ab–3b2

14. (a+4)(a+2) = a2+8

Solution:

LHS = (a+4)(a+2) = a(a+2)+4(a+2)

= a2+2a+4a+8

= a2+6a+8

≠ a2+8 = RHS

The correct statement is (a+4)(a+2) = a2+6a+8

15. (a–4)(a–2) = a2–8

Solution:

LHS = (a–4)(a–2) = a(a–2)–4(a–2)

= a2–2a–4a+8

= a2–6a+8

≠ a2-8 = RHS

The correct statement is (a–4)(a–2) = a2–6a+8

16. 3x2/3x2 = 0

Solution:

LHS = 3x2/3x2 = 1 ≠ 0 = RHS

The correct statement is 3x2/3x2 = 1

17. (3x2+1)/3x2 = 1 + 1 = 2

Solution:

LHS = (3x2+1)/3x2 = (3x2/3x2)+(1/3x2) = 1+(1/3x2) ≠ 2 = RHS

The correct statement is (3x2+1)/3x2 = 1+(1/3x2)

18. 3x/(3x+2) = ½

Solution:

LHS = 3x/(3x+2) ≠ 1/2 = RHS

The correct statement is 3x/(3x+2) = 3x/(3x+2)

19. 3/(4x+3) = 1/4x

Solution:

LHS = 3/(4x+3) ≠ 1/4x

The correct statement is 3/(4x+3) = 3/(4x+3)

20. (4x+5)/4x = 5

Solution:

LHS = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠ 5 = RHS

The correct statement is (4x+5)/4x = 1 + (5/4x) Solution:

LHS = (7x+5)/5 = (7x/5)+ 5/5 = (7x/5)+1 ≠ 7x = RHS

The correct statement is (7x+5)/5 = (7x/5) +1

 Also Access CBSE Notes for Class 8 Maths Chapter 14

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

Class 8 NCERT exercise wise questions and answers will help students frame a perfect solution in the Maths Exam. These exercise questions can provide students with a topic wise preparation strategy. Some of the important topics introduced in Class 8 NCERT Solutions Maths are Factorisation, Factors of natural numbers, Factors of algebraic expressions and Division of Algebraic Expressions.

### NCERT Solutions For Class 8 Maths Chapter 14 Exercises:

Get a detailed solution for all the questions listed under the below exercises:

Exercise 14.1 Solutions: 3 Questions (Short answer type)

Exercise 14.2 Solutions: 5 Questions (Short answer type)

Exercise 14.3 Solutions: 5 Questions (Short answer type)

Exercise 14.4 Solutions: 21 Questions (Short answer type)

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

NCERT Class 8 Maths Chapter 14, deals primarily with the factorisation of the numbers and algebraic expressions using algebraic identities. Students will also learn about, Division of Algebraic Expressions, Division of a monomial by another monomial, Division of a polynomial by a monomial and Division of Polynomial by Polynomial.

The main topics covered in this chapter include:

 Exercise Topic 14.1 Introduction 14.2 What is Factorisation? 14.3 Division of Algebraic Expressions 14.4 Division of Algebraic Expressions 14.5 Can you Find the Error?

Key Features of NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

1. These NCERT solutions help students in understanding the concepts clearly.
2. Simple and easy language is used to explain the topics.
3. All concepts have been explained in detail.
4. Subject experts have consolidated all exercises questions at one place for practice.
5. NCERT Solutions are helpful for the preparation of competitive exams.
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