# NCERT Solutions For Class 8 Maths Chapter 14 Factorisation

## Class 8 Maths Chapter 14 Factorisation

NCERT Solutions For Class 8 Maths Chapter 14 Factorisation, (Maths) exam are Students are taught thru NCERT books in some of state board and CBSE Schools.  As the chapter involves an end, there is an exercise provided to assist students prepare for evaluation.  Students need to clear up those exercises very well because the questions withinside the very last asked from those.

Sometimes, students get stuck withinside the exercises and are not able to clear up all of the questions.  To assist students, solve all of the questions and maintain their studies without a doubt, we have provided step by step NCERT Solutions for the students for all classes.  These answers will similarly help students in scoring better marks with the assist of properly illustrated Solutions as a way to similarly assist the students and answering the questions right.

## 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, 3xand 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 ab = 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 16x, -4x2and 32x

16 x= 2×2×2×2×x×x×x

– 4x2 = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x, – 4xand 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

10xy2 = 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

10a= 2×5×a×a

– 15b= -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+xyz= xyz(x+y+z)

(x) ax2y+bxy2+cxyz

ax2y = a×x×x×y

bxy= b×x×y×y

cxyz = c×x×y×z

Common factors of a x2y ,bxyand 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)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+an= 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)÷ 4xyz2

(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)
First 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

Find and correct the errors in the following mathematical statements.

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

Benefits of NCERT Solutions for Class 8

NCERT Solutions for Class 8 contains extremely important points, and for each chapter, each concept has been simplified to make it easier to remember and increase your chances of achieving excellent exam results. Exam Preparation References Here are some tips on how these Solutions can help you prepare for the exam.

1. This helps students solve many of the problems in each chapter and encourages them to make their concepts more meaningful.
2. NCERT Solutions for Class 8 encourage you to update your knowledge and refine your concepts so that you can get good results in the exam.
3. These NCERT Solutions For Class 8 are the best exam materials, allowing you to learn more about your week and your strengths. To get good results in the exam, it is important to overcome your weaknesses.
4. Most of the questions in the exam are formulated in a similar way to NCERT textbooks. Therefore, students should review the Solutions in each chapter in order to better understand the topic.
5. It is free of cost.

Tips & Strategies for Class 8 Exam Preparation

1. Plan your course and syllabus and make time for revision
2. Please refer to the NCERT Solutions available on the cbsestudyguru website to clarify your concepts every time you prepare for the exam.
3. Use the cbsestudyguru learning app to start learning to successfully pass the exam. Provide complete teaching materials, including resolved and unresolved tasks.
4. It is important to clear all your doubts before the exam with your teachers or Alex (an Al study Bot).
5. When you read or study a chapter, write down algorithm formulas, theorems, etc., and review them quickly before the exam.
6. Practice an ample number of question papers to make your concepts stronger.
7. Take rest and a proper meal.  Don’t stress too much.

Why opt for cbsestudyguru NCERT Solutions for Class 8 ?

• cbsestudyguru provide NCERT Solutions for all subjects at your fingertips.
• These Solutions are designed by subject matter experts and provide Solutions to every NCERT textbook questions.
• cbsestudyguru especially focuses on making learning interactive, effective and for all classes.
• We provide free NCERT Solutions for class 8 and all other classes.