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Chapter 11

Factorisation — Multiple Choice Questions

Class - 8 ML Aggarwal Understanding ICSE Mathematics

Objective Type Questions - Multiple Choice Questions

Question 3

HCF of 6abc6abc, 24ab224ab^2, 12a2b12a^2b is

  1. 6ab6ab

  2. 6ab26ab^2

  3. 6a2b6a^2b

  4. 6abc6abc

Answer

HCF of numerical coefficients = HCF of 6, 24 and 12 = 6

HCF of literal coefficients = HCF of abcabc, ab2ab^2 and a2ba^2b

= product of each common literal raised to the lowest power

= ab

HCF of the given monomials = 6 × abab = 6ab6ab

Hence, option 1 is the correct option.

Question 4

Factors of 12a2b+15ab212a^2b + 15ab^2 are

  1. 3a(4ab+5b2)3a(4ab + 5b^2)

  2. 3ab(4a+5b)3ab(4a + 5b)

  3. 3b(4a2+5ab)3b(4a^2 + 5ab)

  4. none of these

Answer

HCF of 12a2b12a^2b and 15ab215ab^2 is 3ab3ab

Divide each term by 3ab3ab and keep 3ab3ab outside the bracket.

⇒12a2b+15ab2⇒3ab(4a+5b).\Rightarrow 12a^2b + 15ab^2 \\[1em] \Rightarrow 3ab(4a + 5b).

Hence, option 2 is the correct option.

Question 5

Factors of 6xy−4y+6−9x6xy - 4y + 6 - 9x are

  1. (3y−2)(2x−3)(3y - 2)(2x - 3)

  2. (3x−2)(2y−3)(3x - 2)(2y - 3)

  3. (2y−3)(2−3x)(2y - 3)(2 - 3x)

  4. none of these

Answer

Factorising,

⇒6xy−4y+6−9x⇒(6xy−4y)−(9x−6)⇒2y(3x−2)−3(3x−2)⇒(3x−2)(2y−3).\Rightarrow 6xy - 4y + 6 - 9x \\[1em] \Rightarrow (6xy - 4y) - (9x - 6) \\[1em] \Rightarrow 2y(3x - 2) - 3(3x - 2) \\[1em] \Rightarrow (3x - 2)(2y - 3).

Hence, option 2 is the correct option.

Question 6

Factors of 49p3q−36pq49p^3q - 36pq are

  1. p(7p+6q)(7p−6q)p(7p + 6q)(7p - 6q)

  2. q(7p−6)(7p+6)q(7p - 6)(7p + 6)

  3. pq(7p+6)(7p−6)pq(7p + 6)(7p - 6)

  4. none of these

Answer

Factorising,

⇒ 49p3q−36pq49p^3q - 36pq

⇒ pq(49p2−36)pq(49p^2 - 36)

⇒ pq((7p)2−(6)2)pq((7p)^2 - (6)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ pq(7p+6)(7p−6)pq(7p + 6)(7p - 6)

Hence, option 3 is the correct option.

Question 7

Factors of y(y−z)+9(z−y)y(y - z) + 9(z - y) are

  1. (y−z)(y+9)(y - z)(y + 9)

  2. (z−y)(y+9)(z - y)(y + 9)

  3. (y−z)(y−9)(y - z)(y - 9)

  4. none of these

Answer

Factorising,

⇒y(y−z)+9(z−y)⇒y(y−z)−9(y−z)⇒(y−z)(y−9).\Rightarrow y(y - z) + 9(z - y) \\[1em] \Rightarrow y(y - z) - 9(y - z) \\[1em] \Rightarrow (y - z)(y - 9).

Hence, option 3 is the correct option.

Question 8

Factors of (lm+l)+m+1(lm + l) + m + 1 are

  1. (lm+1)(m+l)(lm + 1)(m + l)

  2. (lm+m)(l+1)(lm + m)(l + 1)

  3. l(m+1)l(m + 1)

  4. (l+1)(m+1)(l + 1)(m + 1)

Answer

Factorising,

⇒(lm+l)+m+1⇒l(m+1)+1(m+1)⇒(m+1)(l+1).\Rightarrow (lm + l) + m + 1 \\[1em] \Rightarrow l(m + 1) + 1(m + 1) \\[1em] \Rightarrow (m + 1)(l + 1).

Hence, option 4 is the correct option.

Question 9

Factors of z2−4z−12z^2 - 4z - 12 are

  1. (z+6)(z−2)(z + 6)(z - 2)

  2. (z−6)(z+2)(z - 6)(z + 2)

  3. (z−6)(z−2)(z - 6)(z - 2)

  4. (z+6)(z+2)(z + 6)(z + 2)

Answer

We want to find two integers whose sum is −4 and product is −12.

By trial, we see that (−6) + 2 = −4 and (−6) × 2 = −12

⇒z2−4z−12⇒z2−6z+2z−12⇒z(z−6)+2(z−6)⇒(z−6)(z+2).\Rightarrow z^2 - 4z - 12 \\[1em] \Rightarrow z^2 - 6z + 2z - 12 \\[1em] \Rightarrow z(z - 6) + 2(z - 6) \\[1em] \Rightarrow (z - 6)(z + 2).

Hence, option 2 is the correct option.

Question 10

Factors of 63a2−112b263a^2 - 112b^2 are

  1. 63(a−2b)(a+2b)63(a - 2b)(a + 2b)

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

  3. 7(3a+4b)(3a−4b)7(3a + 4b)(3a - 4b)

  4. none of these

Answer

Factorising,

⇒ 63a2−112b263a^2 - 112b^2

⇒ 7(9a2−16b2)7(9a^2 - 16b^2)

⇒ 7((3a)2−(4b)2)7((3a)^2 - (4b)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ 7(3a+4b)(3a−4b)7(3a + 4b)(3a - 4b)

Hence, option 3 is the correct option.

Question 11

Factors of p4−81p^4 - 81 are

  1. (p2−9)(p2+9)(p^2 - 9)(p^2 + 9)

  2. (p+3)2(p−3)2(p + 3)^2(p - 3)^2

  3. (p+3)(p−3)(p2+9)(p + 3)(p - 3)(p^2 + 9)

  4. none of these

Answer

Factorising,

⇒ p4−81p^4 - 81

⇒ (p2)2−(9)2(p^2)^2 - (9)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (p2+9)(p2−9)(p^2 + 9)(p^2 - 9)

⇒ (p2+9)((p)2−(3)2)(p^2 + 9)((p)^2 - (3)^2)

⇒ (p+3)(p−3)(p2+9)(p + 3)(p - 3)(p^2 + 9)

Hence, option 3 is the correct option.

Question 12

Factors of 3x2+7x−63x^2 + 7x - 6 are

  1. (3x−2)(x+3)(3x - 2)(x + 3)

  2. (3x+2)(x−3)(3x + 2)(x - 3)

  3. (3x−2)(x−3)(3x - 2)(x - 3)

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

Answer

We want to find two integers whose sum is 7 and product is 3 × (−6) i.e. −18.

By trial, we see that 9 + (−2) = 7 and 9 × (−2) = −18

⇒3x2+7x−6⇒3x2+9x−2x−6⇒3x(x+3)−2(x+3)⇒(x+3)(3x−2).\Rightarrow 3x^2 + 7x - 6 \\[1em] \Rightarrow 3x^2 + 9x - 2x - 6 \\[1em] \Rightarrow 3x(x + 3) - 2(x + 3) \\[1em] \Rightarrow (x + 3)(3x - 2).

Hence, option 1 is the correct option.

Question 13

Factors of 16x2+40x+2516x^2 + 40x + 25 are

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

  2. (4x+5)(4x−5)(4x + 5)(4x - 5)

  3. (4x+5)(4x+8)(4x + 5)(4x + 8)

  4. none of these

Answer

Factorising,

⇒ 16x2+40x+2516x^2 + 40x + 25

⇒ (4x)2+2×4x×5+(5)2(4x)^2 + 2 \times 4x \times 5 + (5)^2

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

⇒ (4x+5)2(4x + 5)^2

⇒ (4x+5)(4x+5)(4x + 5)(4x + 5)

Hence, option 1 is the correct option.

Question 14

Factors of x2−4xy+4y2x^2 - 4xy + 4y^2 are

  1. (x−2y)(x+2y)(x - 2y)(x + 2y)

  2. (x−2y)(x−2y)(x - 2y)(x - 2y)

  3. (x+2y)(x+2y)(x + 2y)(x + 2y)

  4. none of these

Answer

Factorising,

⇒ x2−4xy+4y2x^2 - 4xy + 4y^2

⇒ (x)2−2×x×2y+(2y)2(x)^2 - 2 \times x \times 2y + (2y)^2

Using the identity,

[∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

⇒ (x−2y)2(x - 2y)^2

⇒ (x−2y)(x−2y)(x - 2y)(x - 2y)

Hence, option 2 is the correct option.

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