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Chapter 5

Factorisation — Exercise 5(C)

Class - 9 Concise Mathematics Selina

Exercise 5(C)

Question 1(a)

x4 - 1 in the form of factors is :

  1. (x + 1)(x - 1)(x2 - 1)

  2. (x - 1)(x + 1)(x2 + 1)

  3. (x - 1)(x - 2)(x2 + 2)

  4. (x + 1)(x + 2)(x2 - 1)

Answer

Given,

   x4 - 1

= (x2)2 - 12

= (x2 - 1)(x2 + 1)

= (x - 1)(x + 1)(x2 + 1).

Hence, Option 2 is the correct option.

Question 1(b)

2a2 - 18 in the form of factors is :

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

  2. 2(a + 1)(a - 9)

  3. 2(a - 1)(a - 9)

  4. (2a - 1)(a - 9)

Answer

Given,

   2a2 - 18

= 2(a2 - 9)

= 2[(a)2 - (3)2]

= 2(a + 3)(a - 3).

Hence, Option 1 is the correct option.

Question 1(c)

27x3 - 48x in the form of factors is :

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

  2. 3x(3x + 4)(3x + 4)

  3. 3(3x2 - 4x)(3x + 4)

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

Answer

Given,

   27x3 - 48x

= 3x(9x2 - 16)

= 3x[(3x)2 - (4)2]

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

Hence, Option 4 is the correct option.

Question 1(d)

x2 - (x - 4y)2 in the form of factors is :

  1. 8y(x - 2y)

  2. 8y(x + 2y)

  3. 4y(x + 2y)

  4. 4y(x - 2y)

Answer

Given,

   x2 - (x - 4y)2

= [x + (x - 4y)][x - (x - 4y)]

= (2x - 4y)[x - x + 4y]

= 4y(2x - 4y)

= 8y(x - 2y).

Hence, Option 1 is the correct option.

Question 1(e)

9x2+19x2+19x^2 + \dfrac{1}{9x^2} + 1 in the form of factors is :

  1. (3x+13x+1)(3x+13x−1)\Big(3x + \dfrac{1}{3x} + 1\Big)\Big(3x + \dfrac{1}{3x} - 1\Big)

  2. (3x−13x+1)(3x−13x−1)\Big(3x - \dfrac{1}{3x} + 1\Big)\Big(3x - \dfrac{1}{3x} - 1\Big)

  3. (3x+13x)(3x−13x+1)\Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x} + 1\Big)

  4. (3x−13x)(3x+13x−1)\Big(3x - \dfrac{1}{3x}\Big)\Big(3x + \dfrac{1}{3x} - 1\Big)

Answer

Given,

=9x2+19x2+1=9x2+19x2+2−1=(3x)2+(13x)2+2−1=(3x+13x)2−12=(3x+13x+1)(3x+13x−1).\phantom{=} 9x^2 + \dfrac{1}{9x^2} + 1 \\[1em] = 9x^2 + \dfrac{1}{9x^2} + 2 - 1 \\[1em] = (3x)^2 + \Big(\dfrac{1}{3x}\Big)^2 + 2 - 1 \\[1em] = \Big(3x + \dfrac{1}{3x}\Big)^2 - 1^2 \\[1em] = \Big(3x + \dfrac{1}{3x} + 1\Big)\Big(3x + \dfrac{1}{3x} - 1\Big).

Hence, Option 1 is the correct option.

Question 2

Factorise :

a2 - (2a + 3b)2

Answer

Given,

   a2 - (2a + 3b)2

= (a + 2a + 3b)[a - (2a + 3b)]

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

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

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

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

Question 3

Factorise :

25(2a - b)2 - 81b2

Answer

Given,

   25(2a - b)2 - 81b2

= [5(2a - b)]2 - (9b)2

= [5(2a - b) + 9b][5(2a - b) - 9b]

= (10a - 5b + 9b)(10a - 5b - 9b)

= (10a + 4b)(10a - 14b)

= 2(5a + 2b) × 2(5a - 7b)

= 4(5a + 2b)(5a - 7b).

Hence, 25(2a - b)2 - 81b2 = 4(5a + 2b)(5a - 7b).

Question 4

Factorise :

50a3 - 2a

Answer

Given,

   50a3 - 2a

= 2a[25a2 - 1]

= 2a[(5a)2 - 12]

= 2a(5a + 1)(5a - 1).

Hence, 50a3 - 2a = 2a(5a + 1)(5a - 1).

Question 5

Factorise :

4a2b - 9b3

Answer

Given,

   4a2b - 9b3

= b[4a2 - 9b2]

= b[(2a)2 - (3b)2]

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

Hence, 4a2b - 9b3 = b(2a + 3b)(2a - 3b).

Question 6

Factorise :

9(a - 2)2 - 16(a + 2)2

Answer

Given,

   9(a - 2)2 - 16(a + 2)2

= {[3(a - 2)]2 - [4(a + 2)]2}

= {[3(a - 2) + 4(a + 2)][3(a - 2) - 4(a + 2)]}

= [(3a - 6 + 4a + 8)(3a - 6 - 4a - 8)]

= (7a + 2)(-a - 14)

= -(7a + 2)(a + 14).

Hence, 9(a - 2)2 - 16(a + 2)2 = -(7a + 2)(a + 14).

Question 7

(a + b)3 - a - b

Answer

Given,

   (a + b)3 - a - b

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

= (a + b)[(a + b)2 - 1]

= (a + b)[(a + b)2 - 12]

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

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

Question 8

a(a - 1) - b(b - 1)

Answer

Given,

   a(a - 1) - b(b - 1)

= a2 - a - b2 + b

= a2 - b2 - a + b

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

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

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

Question 9

4a2 - (4b2 + 4bc + c2)

Answer

Given,

   4a2 - (4b2 + 4bc + c2)

= 4a2 - [(2b)2 + 2 × 2b × c + c2]

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

= (2a + 2b + c)[2a - (2b + c)]

= (2a + 2b + c)(2a - 2b - c).

Hence, 4a2 - (4b2 + 4bc + c2) = (2a + 2b + c)(2a - 2b - c).

Question 10

4a2 - 49b2 + 2a - 7b

Answer

Given,

   4a2 - 49b2 + 2a - 7b

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

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

= (2a - 7b)(2a + 7b + 1).

Hence, 4a2 - 49b2 + 2a - 7b = (2a - 7b)(2a + 7b + 1).

Question 11

4a2 - 12a + 9 - 49b2

Answer

Given,

   4a2 - 12a + 9 - 49b2

= (2a)2 - 2 × 2a × 3 + (3)2 - (7b)2

= (2a - 3)2 - (7b)2

= (2a - 3 + 7b)(2a - 3 - 7b).

Hence, 4a2 - 12a + 9 - 49b2 = (2a - 3 + 7b)(2a - 3 - 7b).

Question 12

4xy - x2 - 4y2 + z2

Answer

Given,

   4xy - x2 - 4y2 + z2

= -[x2 + 4y2 - 4xy - z2]

= -[(x)2 + (2y)2 - 2 × x × 2y - (z)2]

= -[(x - 2y)2 - (z)2]

= (z)2 - (x - 2y)2

= (z + x - 2y)(z - x + 2y).

Hence, 4xy - x2 - 4y2 + z2 = (z + x - 2y)(z - x + 2y).

Question 13

a2 + b2 - c2 - d2 + 2ab - 2cd

Answer

Given,

   a2 + b2 - c2 - d2 + 2ab - 2cd

= a2 + b2 + 2ab - c2 - d2 - 2cd

= (a + b)2 - (c2 + d2 + 2cd)

= (a + b)2 - (c + d)2

= (a + b + c + d)(a + b - c - d).

Hence, a2 + b2 - c2 - d2 + 2ab - 2cd = (a + b + c + d)(a + b - c - d).

Question 14

4x2 - 12ax - y2 - z2 - 2yz + 9a2

Answer

Given,

   4x2 - 12ax - y2 - z2 - 2yz + 9a2

= 4x2 - 12ax + 9a2 - y2 - z2 - 2yz

= (2x)2 - 2 × 2x × 3a + (3a)2 - (y2 + z2 + 2yz)

= (2x - 3a)2 - (y + z)2

= (2x - 3a + y + z)[2x - 3a - (y + z)]

= (2x - 3a + y + z)(2x - 3a - y - z).

Hence, 4x2 - 12ax - y2 - z2 - 2yz + 9a2 = (2x - 3a + y + z)(2x - 3a - y - z).

Question 15

(a2 - 1)(b2 - 1) + 4ab

Answer

Given,

   (a2 - 1)(b2 - 1) + 4ab

= a2b2 - a2 - b2 + 1 + 4ab

= a2b2 - a2 - b2 + 1 + 2ab + 2ab

= a2b2 + 2ab + 1 - a2 - b2 + 2ab

= (ab + 1)2 - (a2 + b2 - 2ab)

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

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

Hence, (a2 - 1)(b2 - 1) + 4ab = (ab + 1 + a - b)(ab + 1 - a + b).

Question 16

x4 + x2 + 1

Answer

Given,

   x4 + x2 + 1

Adding and subtracting x2 in the polynomial,

⇒ x4 + x2 + 1 + x2 - x2

= x4 + 2x2 + 1 - x2

= (x2)2 + 2 × x2 × 1 + (1)2 - (x)2

= (x2 + 1)2 - (x)2

= (x2 + 1 + x)(x2 + 1 - x).

Hence, x4 + x2 + 1 = (x2 + 1 + x)(x2 + 1 - x).

Question 17

(a2 + b2 - 4c2)2 - 4a2b2

Answer

Given,

   (a2 + b2 - 4c2)2 - 4a2b2

= (a2 + b2 - 4c2)2 - (2ab)2

= (a2 + b2 - 4c2 + 2ab)(a2 + b2 - 4c2 - 2ab)

= (a2 + b2 + 2ab - 4c2)(a2 + b2- 2ab - 4c2)

= [(a + b)2 - (2c)2][(a - b)2 - (2c)2]

= (a + b + 2c)(a + b - 2c)(a - b + 2c)(a - b - 2c).

Hence, (a2 + b2 - 4c2)2 - 4a2b2 = (a + b + 2c)(a + b - 2c)(a - b + 2c)(a - b - 2c).

Question 18

(x2 + 4y2 - 9z2)2 - 16x2y2

Answer

Given,

   (x2 + 4y2 - 9z2)2 - 16x2y2

= (x2 + 4y2 - 9z2)2 - (4xy)2

= (x2 + 4y2 - 9z2 + 4xy)(x2 + 4y2 - 9z2 - 4xy)

= [x2 + (2y)2 + 4xy - 9z2][x2 + (2y)2 - 4xy - 9z2]

= [(x + 2y)2 - (3z)2][(x - 2y)2 - (3z)2]

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

Hence, (x2 + 4y2 - 9z2)2 - 16x2y2 = (x + 2y + 3z)(x + 2y - 3z)(x - 2y + 3z)(x - 2y - 3z).

Question 19

(a + b)2 - a2 + b2

Answer

Given,

   (a + b)2 - a2 + b2

= (a + b)2 - (a2 - b2)

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

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

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

= 2b(a + b).

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

Question 20

a2 - b2 - (a + b)2

Answer

Given,

   a2 - b2 - (a + b)2

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

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

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

= -2b(a + b).

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

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