KnowledgeBoat Logo
|
OPEN IN APP

Chapter 4

Factorisation — Exercise 4(A)

Class - 9 RS Aggarwal Mathematics Solutions



Exercise 4A

Question 1

Factorize :

5x2 - 20xy

Answer

Given,

⇒ 5x2 - 20xy

⇒ 5x(x - 4y)

Hence, 5x2 - 20xy = 5x(x - 4y).

Question 2

Factorize :

18a2b - 24abc

Answer

Given,

⇒ 18a2b - 24abc

⇒ 6ab(3a - 4c)

Hence, 18a2b - 24abc = 6ab(3a - 4c).

Question 3

Factorize :

27x3y3 - 45x4y2

Answer

Given,

⇒ 27x3y3 - 45x4y2

⇒ 9x3y2(3y - 5x)

Hence, 27x3y3 - 45x4y2 = 9x3y2(3y - 5x).

Question 4

Factorize :

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

Answer

Given,

⇒ 5a(b + c) - 7b(b + c)

⇒ (b + c)(5a - 7b)

Hence, 5a(b + c) - 7b(b + c) = (b + c)(5a - 7b).

Question 5

Factorize :

2x(p2 + q2) + 4y(p2 + q2)

Answer

Given,

⇒ 2x(p2 + q2) + 4y(p2 + q2)

⇒ (2x + 4y)(p2 + q2)

⇒ 2(x + 2y)(p2 + q2)

Hence, 2x(p2 + q2) + 4y(p2 + q2) = 2(x + 2y)(p2 + q2).

Question 6

Factorize :

x(a - 5) + y(5 - a)

Answer

Given,

⇒ x(a - 5) + y(5 - a)

⇒ x(a - 5) - y(a - 5)

⇒ (a - 5)(x - y)

Hence, x(a - 5) + y(5 - a) = (a - 5)(x - y).

Question 7

Factorize :

4(x + y) - 6(x + y)2

Answer

Given,

⇒ 4(x + y) - 6(x + y)2

⇒ (x + y)[4 - 6(x + y)]

⇒ (x + y).2.[2 - 3(x + y)]

⇒ 2(x + y)(2 - 3x - 3y)

Hence, 4(x + y) - 6(x + y)2 = 2(x + y)(2 - 3x - 3y).

Question 8

Factorize :

6(2a + 3b)2 - 8(2a + 3b)

Answer

Given,

⇒ 6(2a + 3b)2 - 8(2a + 3b)

⇒ (2a + 3b)[6(2a + 3b) - 8]

⇒ (2a + 3b).2.[3(2a + 3b) - 4]

⇒ 2(2a + 3b)(6a + 9b - 4).

Hence, 6(2a + 3b)2 - 8(2a + 3b) = 2(2a + 3b)(6a + 9b - 4).

Question 9

Factorize :

x(x + y)3 - 3x2y(x + y)

Answer

Given,

⇒ x(x + y)3 - 3x2y(x + y)

⇒ x(x + y)[(x + y)2 - 3xy]

⇒ x(x + y)[x2 + y2 + 2xy - 3xy]

⇒ x(x + y)(x2 + y2 - xy).

Hence, x(x + y)3 - 3x2y(x + y) = x(x + y)(x2 + y2 - xy).

Question 10

Factorize :

a3 + 2a2 + 5a + 10

Answer

Given,

⇒ a3 + 2a2 + 5a + 10

⇒ a2(a + 2) + 5(a + 2)

⇒ (a + 2)(a2 + 5).

Hence, a3 + 2a2 + 5a + 10 = (a + 2) (a2 + 5).

Question 11

Factorize :

x2 + xy - 2xz - 2yz

Answer

Given,

⇒ x2 + xy - 2xz - 2yz

⇒ x(x + y) -2z (x + y)

⇒ (x + y)(x - 2z)

Hence, x2 + xy - 2xz - 2yz = (x + y)(x - 2z).

Question 12

Factorize :

a3b - a2b + 5ab - 5b

Answer

Given,

⇒ a3b - a2b + 5ab - 5b

⇒ a2b(a - 1) + 5b (a - 1)

⇒ (a2b + 5b)(a - 1)

⇒ b(a2 + 5)(a - 1).

Hence, a3b - a2b + 5ab - 5b = b(a2 + 5)(a - 1).

Question 13

Factorize :

x2 + y - xy - x

Answer

Given,

⇒ x2 + y - xy - x

⇒ x2 - x - xy + y

⇒ x(x - 1) - y(x - 1)

⇒ (x - 1)(x - y).

Hence, x2 + y - xy - x = (x - 1)(x - y).

Question 14

Factorize :

a(a + b - c) - bc

Answer

Given,

⇒ a(a + b - c) - bc

⇒ a2 + ab - ac - bc

⇒ a(a + b) - c(a + b)

⇒ (a - c)(a + b)

Hence, a(a + b - c) - bc = (a - c)(a + b).

Question 15

Factorize :

(4a - 1)2 - 8a + 2

Answer

Given,

⇒ (4a - 1)2 - 8a + 2

⇒ (4a - 1)2 - 8a + 2

⇒ 16a2 - 8a + 1 - 8a + 2

⇒ 16a2 -16a + 3

⇒ 16a2 -12a - 4a + 3

⇒ 4a(4a - 3) - 1(4a - 3)

⇒ (4a - 1)(4a - 3).

Hence, (4a - 1)2 - 8a + 2 = (4a - 1)(4a - 3).

Question 16

Factorize :

8 - 4a - 2a3 + a4

Answer

Given,

⇒ 8 - 4a - 2a3 + a4

⇒ 4(2 - a) - a3(2 - a)

⇒ (2 - a)(4 - a3)

Hence, 8 - 4a - 2a3 + a4 = (2 - a)(4 - a3).

Question 17

Factorize :

2a2 + bc - 2ab - ac

Answer

Given,

⇒ 2a2 + bc - 2ab - ac

⇒ 2a2 - 2ab - ac + bc

⇒ 2a(a - b) - c (a - b)

⇒ (2a - c)(a - b).

Hence, 2a2 + bc - 2ab - ac = (2a - c)(a - b).

Question 18

Factorize :

a(a - 2b - c) + 2bc

Answer

Given,

⇒ a(a - 2b - c) + 2bc

⇒ a2 - 2ab - ac + 2bc

⇒ a2 - ac - 2ab + 2bc

⇒ a(a - c) - 2b(a - c)

⇒ (a - c)(a - 2b)

Hence, a(a - 2b - c) + 2bc = (a - c)(a - 2b).

Question 19

Factorize :

x2 - (a + b)x + ab

Answer

Given,

⇒ x2 - (a + b)x + ab

⇒ x2 - ax - bx + ab

⇒ x(x - a) - b(x - a)

⇒ (x - a)(x - b).

Hence, x2 - (a + b)x + ab = (x - a)(x - b).

Question 20

Factorize :

3ax - 6ay - 8by + 4bx

Answer

Given,

⇒ 3ax - 6ay - 8by + 4bx

⇒ 3ax + 4bx - 6ay - 8by

⇒ x(3a + 4b) - 2y(3a + 4b)

⇒ (3a + 4b)(x - 2y)

Hence, 3ax - 6ay - 8by + 4bx = (3a + 4b)(x - 2y).

Question 21

Factorize :

ab(x2 + y2) - xy(a2 + b2)

Answer

Given,

⇒ ab(x2 + y2) - xy(a2 + b2)

⇒ abx2 + aby2 - xya2 - xyb2

⇒ abx2 - xya2 - xyb2 + aby2

⇒ ax(bx - ay) - by(bx - ay)

⇒ (bx - ay)(ax - by).

Hence, ab(x2 + y2) - xy(a2 + b2) = (bx - ay)(ax - by).

Question 22

Factorize :

ab(x2 + 1) + x(a2 + b2)

Answer

Given,

⇒ ab(x2 + 1) + x(a2 + b2)

⇒ abx2 + ab + xa2 + xb2

⇒ abx2 + xa2 + xb2 + ab

⇒ ax(bx + a) + b(bx + a)

⇒ (ax + b)(bx + a).

Hence, ab(x2 + 1) + x(a2 + b2) = (ax + b)(bx + a).

Question 23

Factorize :

a3 + ab(1 - 2a) - 2b2

Answer

Given,

⇒ a3 + ab(1 - 2a) - 2b2

⇒ a3 + ab - 2a2b - 2b2

⇒ a3 - 2a2b + ab - 2b2

⇒ a2(a - 2b) + b(a - 2b)

⇒ (a - 2b)(a2 + b).

Hence, a3 + ab(1 - 2a) - 2b2 = (a - 2b)(a2 + b).

Question 24

Factorize :

x2+1x223x+3xx^2 + \dfrac{1}{x^2} - 2 - 3x + \dfrac{3}{x}

Answer

Given,

x2+1x223x+3x(x1x)23(x1x)(x1x)[(x1x)3](x1x)(x1x3)\Rightarrow x^2 + \dfrac{1}{x^2} - 2 - 3x + \dfrac{3}{x} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 - 3\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)\Big[\Big(x - \dfrac{1}{x}\Big) - 3\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 3\Big)

Hence, x2+1x223x+3x=(x1x)(x1x3)x^2 + \dfrac{1}{x^2} - 2 - 3x + \dfrac{3}{x} = \Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 3\Big).

PrevNext