KnowledgeBoat Logo
|
OPEN IN APP

Chapter 4

Factorisation — Multiple Choice Questions

Class - 9 RS Aggarwal Mathematics Solutions



Multiple Choice Questions

Question 1

b2 - ac - bc + ab =

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

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

  3. (a - b)(b + c)

  4. (a - b)(b - c)

Answer

Given,

⇒ b2 - ac - bc + ab

⇒ b2 + ab - ac - bc

⇒ b(b + a) - c(a + b)

⇒ (a + b)(b - c).

Hence, option 1 is correct option.

Question 2

The value of (1 + x)2(1 + y2) - (1 + x2)(1 + y)2 is:

  1. 2(x - y)(1 + xy)

  2. (x - y)(1 - xy)

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

  4. 2(x + y)(1 - xy)

Answer

Given,

⇒ (1 + x)2(1 + y2) - (1 + x2)(1 + y)2

⇒ (1 + 2x + x2)(1 + y2) - (1 + x2)(1 + 2y + y2)

⇒ (1 + y2 + 2x + 2xy2 + x2 + x2y2) - (1 + 2y + y2 + x2 + 2x2y + x2y2)

⇒ (1 + y2 + 2x + 2xy2 + x2 + x2y2 - 1 - 2y - y2 - x2 - 2x2y - x2y2)

⇒ 2x - 2y - 2x2y + 2xy2

⇒ 2(x - y) - 2xy(x - y)

⇒ (2 - 2xy)(x - y)

⇒ 2(1 - xy)(x - y).

Hence, option 3 is correct option.

Question 3

x4 - y4 = ............... (x - y)(x2 + y2)

  1. 0

  2. 1

  3. x − y

  4. x + y

Answer

Given,

⇒ x4 - y4

⇒ (x2)2 - (y2)2

⇒ (x2 - y2)(x2 + y2)

⇒ [(x)2 - (y)2](x2 + y2)

⇒ (x + y)(x - y)(x2 + y2)

Hence, option 4 is correct option.

Question 4

16(2a − b)2 − 9(a + b)2 =

  1. (5a − 7b)(11a − b)

  2. (5a + 7b)(11a − b)

  3. (5a − 7b)(11a + b)

  4. (5a + 7b)(11a + b)

Answer

Given,

⇒ [4(2a − b)]2 − [3(a + b)]2

⇒ [4(2a − b) − 3(a + b)][4(2a − b) + 3(a + b)]

⇒ (8a − 4b − 3a - 3b)(8a − 4b + 3a + 3b)

⇒ (5a − 7b)(11a − b).

Hence, option 1 is correct option.

Question 5

Factorization of 3a(3a + 2c) − 4b(b + c) is:

  1. (3a + 2b)(3a + 2b + 2c)

  2. (3a − 2b)(3a − 2b + 2c)

  3. (3a − 2b)(3a + 2b + 2c)

  4. (3a − 2b)(3a + 2b − 2c)

Answer

Given,

⇒ 3a(3a + 2c) − 4b(b + c)

⇒ 9a2 + 6ac − 4b2 - 4cb

⇒ (3a)2 − (2b)2 + 6ac - 4cb

⇒ (3a + 2b)(3a - 2b) + 2c(3a - 2b)

⇒ (3a - 2b)(3a + 2b + 2c).

Hence, option 3 is correct option.

Question 6

12a2 - 27b4

  1. 3(2a + 3b2)(2a - b2)

  2. (2a + 3b2)(2a - 3b2)

  3. 3(2a + 3b2)(a - 3b2)

  4. 3(2a + 3b2)(2a - 3b2)

Answer

Given,

⇒ 3(4a2 - 9b4)

⇒ 3[(2a)2 - (3b2)2]

⇒ 3(2a + 3b2)(2a - 3b2)

Hence, option 4 is correct option.

Question 7

3x2+11x+63=\sqrt{3}x^2 + 11x + 6\sqrt{3} =

  1. (3x+2)(x33)(\sqrt{3}x + 2)(x - 3\sqrt{3})

  2. (3x+2)(x+33)(\sqrt{3}x + 2)(x + 3\sqrt{3})

  3. (3x2)(x33)(\sqrt{3}x - 2)(x - 3\sqrt{3})

  4. (x2)(x+33)(x - 2)(x + 3\sqrt{3})

Answer

Given,

3x2+11x+633x2+9x+2x+633x(x+33)+2(x+33)(3x+2)(x+33)\Rightarrow \sqrt{3}x^2 + 11x + 6\sqrt{3} \\[1em] \Rightarrow \sqrt{3}x^2 + 9x + 2x + 6\sqrt{3} \\[1em] \Rightarrow \sqrt{3}x(x + 3\sqrt{3}) + 2(x + 3\sqrt{3}) \\[1em] \Rightarrow (\sqrt{3}x + 2)(x + 3\sqrt{3})

Hence, option 2 is correct option.

Question 8

1 − x9 = (1 − x)(1 + x + x2) ...............

  1. 1 − x3 + x6

  2. 1 + x3 + x6

  3. 1 − x3 − x6

  4. 1 + x3 − x6

Answer

Given,

⇒ 1 − x9

⇒ (1)3 − (x3)3

By using identity,

a3 - b3 = (a - b)(a2 + ab + b2)

⇒ (1 - x3)(12 + x3(1) + (x3)2)

⇒ [(1)3 - (x)3](1 + x3 + x6)

⇒ (1 - x)(12 + x(1) + x2)(1 + x3 + x6)

⇒ (1 - x)(1 + x + x2)(1 + x3 + x6).

Hence, option 2 is correct option.

Question 9

7 − 12m − 4m2 =

  1. (1 − 2m)(7 + 2m)

  2. (1 + 2m)(7 − 2m)

  3. (1 + 2m)(7 + 2m)

  4. (2m − 1)(7 + 2m)

Answer

Given,

⇒ 7 − 12m − 4m2

⇒ 7 − 14m + 2m − 4m2

⇒ 7(1 − 2m) + 2m(1 − 2m)

⇒ (1 − 2m)(7 + 2m).

Hence, option 1 is correct option.

Question 10

Factorization of p2 − 2p − (q + 1)(q − 1) is:

  1. (p + q − 1)(p − q + 1)

  2. (p − q + 1)(p − q − 1)

  3. (p + q − 1)(p − q − 1)

  4. (p + q + 1)(p − q)

Answer

Given,

⇒ p2 − 2p − (q + 1)(q − 1)

⇒ p2 − 2p − (q2 - 1)

⇒ p2 − 2p − q2 + 1

⇒ p2 − 2p + 1 − q2

⇒ (p - 1)2 − q2

⇒ (p - 1 + q)(p - 1 - q)

⇒ (p + q − 1)(p − q - 1).

Hence, option 3 is correct option.

PrevNext