NCERT Solutions for Class 8 Maths Chapter 14: Factorisation

The NCERT Solutions aim to facilitate students in mastering fundamental concepts and gaining a comprehensive understanding of the diverse question types encountered in CBSE Class 8 Mathematics Examinations.

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Exercise 14.1

Question 1 :

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

 

Answer :

 

Ncert solutions class 8 chapter 14-1

Ncert solutions class 8 chapter 14-2

(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)

 


Question 2 :

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

 

Answer :

Ncert solutions class 8 chapter 14-3


Question 3 :

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

 

Answer :

(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 28p2q2

14pq = 2x7xpxq

28p2q2 = 2x2x7xpxpxqxq

Common factors of 14 pq and 28 p2q2 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

 


Exercise 14.2

Question 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

 

Answer :

(i) a2+8a+16

= a2+2×4×a+42

= (a+4)2

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

(ii) p2–10p+25

= p2-2×5×p+52

= (p-5)2

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

(iii) 25m2+30m+9

= (5m)2+2×5m×3+32

= (5m+3)2

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

(iv) 49y2+84yz+36z2

=(7y)2+2×7y×6z+(6z)2

= (7y+6z)2

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

(v) 4x2–8x+4

= (2x)2-2×4x+22

= (2x-2)2

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

(vi) 121b2-88bc+16c2

= (11b)2-2×11b×4c+(4c)2

= (11b-4c)2

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

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

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

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

= l2+m2-2lm

= (l-m)2

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

(viii) a4+2a2b2+b4

= (a2)2+2×a2×b2+(b2)2

= (a2+b2)2

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

 


Question 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

 

Answer :

(i) 4p2–9q2

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

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

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

(ii) 63a2–112b2

= 7(9a2 –16b2)

= 7((3a)2–(4b)2)

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

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

(iii) 49x2–36

= (7x)2 -62

= (7x+6)(7x–6)

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

(iv) 16x5–144x3

= 16x3(x2–9)

= 16x3(x2–9)

= 16x3(x–3)(x+3)

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

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

= {(l+m)-(l–m)}{(l +m)+(l–m)}

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

(vii) (x2–2xy+y2)–z2

= (x–y)2–z2

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

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using the 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 the identity x2-y2 = (x+y)(x-y) , we have

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

 


Question 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

 

Answer :

(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)

 


Question 4 :

Factorise.

(i) a4–b4

(ii) p4–81

(iii) x4–(y+z) 4

(iv) x4–(x–z) 4

(v) a4–2a2b2+b4

 

Answer :

(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

= (a – b)2 (a + b)2

 


Question 5 :

Factorise the following expressions.

(i) p2+6p+8

(ii) q2–10q+21

(iii) p2+6p–16

 

Answer :

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

(ii) q2–10q+21

We 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 that 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

Question 1 :

Carry out the following divisions.

(i) 28x4 ÷ 56x

(ii) –36y3 ÷ 9y2

(iii) 66pq2r3 ÷ 11qr2

(iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (– 6a6b4)

 

Answer :

(i)28x4 = 2×2×7×x×x×x×x

56x = 2×2×2×7×x

Ncert solutions class 8 chapter 14-4

Ncert solutions class 8 chapter 14-5

Ncert solutions class 8 chapter 14-6

Ncert solutions class 8 chapter 14-7

Ncert solutions class 8 chapter 14-8

 


Question 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

 

Answer :

Ncert solutions class 8 chapter 14-9


Question 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)

 

Answer :

(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

Ncert solutions class 8 chapter 14-10

 

 


Question 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)

 

Answer :

Ncert solutions class 8 chapter 14-11


Question 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)

 

Answer :

(i) (y2+7y+10)÷(y+5)

First, solve the 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: Factorise 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)

Factorising 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 the identity.

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

(vi) 12xy(9x2–16y2) ÷ 4xy(3x+4y)

Factorising 9x2–16y2 , we have

9x2–16y2 = (3x)2–(4y)2 = (3x+4y)(3x-4y) using the 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) =

Ncert solutions class 8 chapter 14-12

 


Exercise 14.4

Question 1 :

1. 4(x–5) = 4x–5

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

3. 2x+3y = 5xy

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

5. 5y+2y+y–7y = 0

6. 3x+2x = 5x2

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

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

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

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

11.(y–3)2 = y2–9

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

13. (2a+3b)(a–b) = 2a2–3b2

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

15. (a–4)(a–2) = a2–8

16. 3x2/3x2 = 0

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

18. 3x/(3x+2) = ½

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

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

Ncert solutions class 8 chapter 14-13

 

Answer :

1.4(x- 5)= 4x – 20 ≠ 4x – 5 = RHS

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

2.LHS = x(3x+2) = 3x2+2x ≠ 3x2+2 = RHS

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

3.LHS= 2x+3y ≠ R. H. S

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

4.LHS = x+2x+3x = 6x ≠ RHS

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

5.LHS = 5y+2y+y–7y = y ≠ RHS

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

6.LHS = 3x+2x = 5x ≠ RHS

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

7.LHS = (2x) 2+4(2x)+7 = 4x2+8x+7 ≠ RHS

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

8.LHS = (2x) 2+5x = 4x2+5x ≠ 9x = RHS

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

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

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

10.(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.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. 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.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.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.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.LHS = 3x2/3x2 = 1 ≠ 0 = RHS

The correct statement is 3x2/3x2 = 1

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

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

18.LHS = 3x/(3x+2) ≠ 1/2 = RHS

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

19.LHS = 3/(4x+3) ≠ 1/4x

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

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

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

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

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

 


Frequently Asked Questions

The NCERT solution for class 8 Chapter 14: Factorisation is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tacking more difficult concepts in their further education.

Yes, the NCERT solution for class 8 Chapter 14: Factorisation is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Factorisation ally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.

You can get all the NCERT solutions for class 8 Maths Chapter 14  from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.

Yes, students must practice all the questions provided in the NCERT solution for class 8 Maths Chapter 14 : Factorisation  as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.

Students can utilize the NCERT solution for class 8 Maths Chapter 14  effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution. Also, you can make Factorisation al notes and jot down the important concepts for your understanding.

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