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Chapter 5

Factorisation — Exercise 5(B)

Class - 9 Concise Mathematics Selina



Exercise 5(B)

Question 1(a)

x2 + 2(5x + 12) in the form of factors is :

  1. (x - 6)(x + 4)

  2. (x - 6)(x - 4)

  3. (x + 6)(x + 4)

  4. (x + 6)(x - 4)

Answer

Given,

   x2 + 2(5x + 12)

= x2 + 10x + 24

= x2 + 6x + 4x + 24

= x(x + 6) + 4(x + 6)

= (x + 4)(x + 6).

Hence, Option 3 is the correct option.

Question 1(b)

x(2x - 5) + 3 in the form of factors is :

  1. (x + 1)(2x - 3)

  2. (x + 1)(2x + 3)

  3. (x - 1)(2x + 3)

  4. (x - 1)(2x - 3)

Answer

Given,

   x(2x - 5) + 3

= 2x2 - 5x + 3

= 2x2 - 3x - 2x + 3

= x(2x - 3) - 1(2x - 3)

= (x - 1)(2x - 3).

Hence, Option 4 is the correct option.

Question 1(c)

6x2 - 2 - 4x in the form of factors is :

  1. 2(x - 1)(3x - 1)

  2. 2(x - 1)(3x + 1)

  3. (2x - 1)(3x - 1)

  4. (6x - 2)(x + 1)

Answer

Given,

   6x2 - 2 - 4x

= 6x2 - 4x - 2

= 6x2 - 6x + 2x - 2

= 6x(x - 1) + 2(x - 1)

= (x - 1)(6x + 2)

= 2(x - 1)(3x + 1).

Hence, Option 2 is the correct option.

Question 1(d)

x(x - 1) - 20 in the form of factors is :

  1. (x - 5)(x + 4)

  2. (x - 5)(x - 4)

  3. (x + 5)(x - 4)

  4. (x + 5)(x + 4)

Answer

Given,

   x(x - 1) - 20

= x2 - x - 20

= x2 - 5x + 4x - 20

= x(x - 5) + 4(x - 5)

= (x - 5)(x + 4).

Hence, Option 1 is the correct option.

Question 1(e)

5 - x(6 - x) in the form of factors is :

  1. (1 - x)(5 + x)

  2. (1 + x)(5 + x)

  3. (1 - x)(5 - x)

  4. (1 + x)(5 - x)

Answer

Given,

   5 - x(6 - x)

= 5 - 6x + x2

= x2 - 6x + 5

= x2 - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 1)(x - 5)

= [-(1 - x).-(5 - x)]

= (1 - x)(5 - x).

Hence, Option 3 is the correct option.

Question 2

Factorise :

1 - 2a - 3a2

Answer

Given,

   1 - 2a - 3a2

= -(3a2 + 2a - 1)

= -[3a2 + 3a - a - 1]

= -[3a(a + 1) - 1(a + 1)]

= -(3a - 1)(a + 1)

= (1 - 3a)(a + 1).

Hence, 1 - 2a - 3a2 = (1 - 3a)(a + 1).

Question 3

Factorise :

24a3 + 37a2 - 5a

Answer

Given,

   24a3 + 37a2 - 5a

= 24a3 + 40a2 - 3a2 - 5a

= 8a2(3a + 5) - a(3a + 5)

= (8a2 - a)(3a + 5)

= a(8a - 1)(3a + 5).

Hence, 24a3 + 37a2 - 5a = a(8a - 1)(3a + 5).

Question 4

Factorise :

3 - a(4 + 7a)

Answer

Given,

   3 - a(4 + 7a)

= 3 - 4a - 7a2

= -7a2 - 4a + 3

= -(7a2 + 4a - 3)

= -[7a2 + 7a - 3a - 3]

= -[7a(a + 1) - 3(a + 1)]

= -[(a + 1)(7a - 3)]

= -[-(a + 1)(3 - 7a)]

= (a + 1)(3 - 7a).

Hence, 3 - a(4 + 7a) = (a + 1)(3 - 7a).

Question 5

Factorise :

(2a + b)2 - 6a - 3b - 4

Answer

Given,

   (2a + b)2 - 6a - 3b - 4

= (2a + b)2 - 3(2a + b) - 4

Substituting (2a + b) = x, we get :

= x2 - 3x - 4

= x2 - 4x + x - 4

= x(x - 4) + 1(x - 4)

= (x - 4)(x + 1)

= (2a + b - 4)(2a + b + 1).

Hence, (2a + b)2 - 6a - 3b - 4 = (2a + b - 4)(2a + b + 1).

Question 6

Factorise :

1 - 2a - 2b - 3(a + b)2

Answer

Given,

   1 - 2a - 2b - 3(a + b)2

= 1 - 2(a + b) - 3(a + b)2

Substituting (a + b) = x, we get :

= 1 - 2x - 3x2

= -3x2 - 2x + 1

= -[3x2 + 2x - 1]

= -[3x2 + 3x - x - 1]

= -[3x(x + 1) - 1(x + 1)]

= -[(x + 1)(3x - 1)]

= -[-(x + 1)(1 - 3x)]

= (x + 1)(1 - 3x)

= (a + b + 1)[1 - 3(a + b)]

= (a + b + 1)(1 - 3a - 3b).

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

Question 7

Factorise :

3a2 - 1 - 2a

Answer

Given,

   3a2 - 1 - 2a

= 3a2 - 2a - 1

= 3a2 - 3a + a - 1

= 3a(a - 1) + 1(a - 1)

= (a - 1)(3a + 1).

Hence, 3a2 - 1 - 2a = (a - 1)(3a + 1).

Question 8

Factorise :

x2 + 3x + 2 + ax + 2a

Answer

Given,

   x2 + 3x + 2 + ax + 2a

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

= x2 + 2x + x + 2 + a(x + 2)

= x(x + 2) + 1(x + 2) + a(x + 2)

= (x + 2)(x + 1 + a).

Hence, x2 + 3x + 2 + ax + 2a = (x + 2)(x + 1 + a).

Question 9

Factorise :

(3x - 2y)2 + 3(3x - 2y) - 10

Answer

Given,

   (3x - 2y)2 + 3(3x - 2y) - 10

Substituting (3x - 2y) = a, we get :

⇒ a2 + 3a - 10

= a2 + 5a - 2a - 10

= a(a + 5) - 2(a + 5)

= (a + 5)(a - 2)

= (3x - 2y + 5)(3x - 2y - 2).

Hence, (3x - 2y)2 + 3(3x - 2y) - 10 = (3x - 2y + 5)(3x - 2y - 2).

Question 10

Factorise :

5 - (3a2 - 2a)(6 - 3a2 + 2a)

Answer

Given,

   5 - (3a2 - 2a)(6 - 3a2 + 2a)

= 5 - [-(3a2 - 2a)(3a2 - 2a - 6)]

= 5 + (3a2 - 2a)(3a2 - 2a - 6)

Substituting 3a2 - 2a = x, we get :

= 5 + x(x - 6)

= 5 + x2 - 6x

= x2 - 6x + 5

= x2 - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 5)(x - 1)

= (3a2 - 2a - 5)(3a2 - 2a - 1)

= [3a2 - 5a + 3a - 5] [3a2 - 3a + a - 1]

= [a(3a - 5) + 1(3a - 5)] [3a(a - 1) + 1(a - 1)]

= (3a - 5)(a + 1)(a - 1)(3a + 1).

Hence, 5 - (3a2 - 2a)(6 - 3a2 + 2a) = (3a - 5)(a + 1)(a - 1)(3a + 1).

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