## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Objective Type Questions

**Mental Maths**

Question 1.

Fill in the blanks:

(i) When an algebraic expression can be written as the product of two or more expressions then each of these expressions is called ……….. of the given expression.

(ii) The process of finding two or more expressions whose product is the given expression is called ………..

(iii) HCF of two or more monomials = (HCF of their ……….. coefficients) × (HCF of their literal coefficients)

(iv) HCB of literal coefficients = product of each common literal raised to the ……….. power.

(v) To factorise the trinomial of the form x^{2} + px + q, we need to find two integers a and b such that a + b = ……….. and ab = ………..

(vi) To factorise the trinomial of the form ax^{2} + bx + c, where a, b and c are integers, we split b into two parts such that ……….. of these parts is b and their is ……….. ac.

Solution:

(i) When an algebraic expression can be written as the product of two or

more expressions then each of these expressions is called factor of the given expression.

(ii) The process of finding two or more expressions

whose product is the given expression is called factorization.

(iii) HCF of two or more monomials

= (HCF of their numerical coefficients) × (HCF of their literal coefficients)

(iv) HCF of literal coefficients

= product of each common literal raised to the lowest power.

(v) To factorise the trinomial of form x^{2} + px + q,

we need to find two integers a and b such that a + b= p and ab = q.

(vi) To factorise the trinomial of the form ax^{2} + bx + c,

where a, b and c are integers, we split b into two parts such that

algebraic sum of these parts is b and their product is ac.

Question 2.

State whether the following statements are true (T) or false (F):

(i) Factorisation is the reverse process of multiplication.

(ii) HCF of two or more polynomials (with integral coefficients) is the smallest common factor of the given polynomials.

(iii) HCF of 6x^{2}y^{2} and 8xy^{3} is 2xy^{2}.

(iv) Factorisation by grouping is possible only if the given polynomial contains an even number of terms.

(v) To factorise the trinomial of the form ax^{2} + bx + c where, a, b, c are integers we want to find two integers A and B such that

A + B = ac and AB = b

(vi) Factors of 4x^{2} – 12x + 9 are (2x – 3) (2x – 3).

Solution:

(i) Factorisation is the reverse process of multiplication. True

(ii) HCF of two or more polynomials (with integral coefficients) is the

smallest common factor of the given polynomials. False

(iii) HCF of 6x^{2}y^{2} and 8xy^{2} is 2xy^{2}. True

(iv) Factorisation by grouping is possible only

if the given polynomial contains an even number of terms. True

(v) To factorise the trinomial of the form ax^{2} + bx + c

where, a, b, c are integers we want to find two integers A and B such that

A + B = ac and AB = b False

Correct :

A + B should be equal to ft and AB = ac

(vi) Factors of

4x^{2} – 12x + 9 are (2x – 3) (2x – 3). True

**Multiple Choice Questions**

**Choose the correct answer from the given four options (3 to 14):**

Question 3.

H.C.F. of 6abc, 24ab^{2}, 12a^{2}b is

(a) 6ab

(b) 6ab^{2}

(c) 6a^{2}b

(d) 6abc

Solution:

H.C.F. of babe, 24ab^{2}, 12a^{2}b

= H.C.F. of 6, 24, 12 × H.C.F. of abc, ab^{2}, a^{2}b

= 6 × a × b = 6ab (a)

Question 4.

Factors of 12a^{2}b + 15ab^{2} are

(a) 3a(4ab + 5b2)

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

(c) 3b(4a^{2} + 5ab)

(d) none of these

Solution:

12a^{2}b + 15 ab^{2 }= 3ab(4a + 5b) (b)

Question 5.

Factors of 6xy – 4y + 6 – 9x are

(a) (3y – 2) (2x – 3)

(b) (3x – 2) (2y – 3)

(c) (2y – 3) (2 – 3x)

(d) none of these

Solution:

6xy – 4y + 6 – 9x

= 6xy – 9x – 4y + 6

= 3x(2y – 3) -2(2y – 3)

= (2y – 3) (3x – 2)

Question 6.

Factors of 49p^{3}q – 36pq are

(a) p(7p + 6q) (7p – 6q)

(b) q(7p – 6) (7p + 6)

(c) pq(7p + 6) (7p – 6)

(d) none of these

Solution:

49p^{2}q – 36pq

= pq(49p^{2} – 36)

=pq[(7p)^{2} – (6)^{2}]

= pq(7p + 6) (7p – 6)

Question 7.

Factors of y(y – z) + 9(z – y) are

(a) (y – z) (y + 9)

(b) (z – y) (y + 9)

(c) (y – z) (y – 9)

(d) none of these

Solution:

y(y – z) + 9(z – y)

= y(y – z) – 9(y – z)

= (y – z) (y – 9) (c)

Question 8.

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

(a) (lm + l )(m + l)

(b) (lm + m)(l + 1)

(c) l(m + 1)

(d) (l + 1)(m + 1)

Solution:

Factors of lm + l + m + 1 are

l(m + 1) + l (m + 1) = (m + 1)(l + 1) (d)

Question 9.

Factors of z^{2} – 4z – 12 are

(a) (z + 6)(z – 2)

(b) (z – 6)(z + 2)

(c) (z – 6)(z – 2)

(d) (z + 6)(z + 2)

Solution:

Factors of z^{2} – 4z – 12

⇒ z^{2} – 6z + 2z – 12

= z(z – 6) + 2(z – b)

= (z – 6)(z + 2) (b)

Question 10.

Factors of 63a^{2} – 112b^{2} are

(a) 63 (a – 2b)(a + 2b)

(b) 7(3a + 2b)(3a – 2b)

(c) 7(3a + 4b)(3a – 4b)

(d) none of these

Solution:

Factors of 63a^{2} – 112b^{2} are

= 7(9a^{2} – 16b^{2})

= 7[(3a)^{2} – (4b)^{2}]

= 7(3a + 4b)(3a – 4b) (c)

Question 11.

Factors of p^{4} – 81 are

(a) (p^{2} – 9)(p^{2} + 9)

(b) (p + 3)^{2} (p – 3)^{2}

(c) (p + 3) (p – 3) (p^{2} + 9)

(d) none of these

Solution:

p^{4} – 81 = (p^{2})^{2} – (9)^{2}

= (p^{2} + 9)(p^{2} – 9)

= (p^{2} + 9){(p)^{2} – (3)^{2}}

= (p^{2} + 9) (p + 3) (p – 3) (c)

Question 12.

Factors of 3x + 7x – 6 are

(a) (3x – 2)(x + 3)

(b) (3x + 2) (x – 3)

(c) (3x – 2)(x – 3)

(d) (3x + 2) (x + 3)

Solution:

3x^{2} + 7x – 6

= 3x^{2} + 9x – 2x – 6

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

= (3x – 2) (x + 3) (a)

Question 13.

Factors of 16x^{2} + 40x + 25 are

(a) (4x + 5)(4x + 5)

(b) (4x + 5)(4x – 5)

(c) (4x + 5)(4x + 8)

(d) none of these

Solution:

16x^{2} + 40x + 25

= (4x)^{2} + 2 × 4x × 5 + (5)^{2}

= (4x + 5)^{2}

= (4x + 5)(4x + 5) (a)

Question 14.

Factors of x^{2} – 4xy + 4y^{2} are

(a) (x – 2y)(x + 2y)

(b) (x-2y)(x-2y)

(c) (x + 2y)(x + 2y)

(d) none of these

Solution:

x^{2} – 4xy + 4y^{2}

= (x)^{2} – 2 × x × 2y + (2y)^{2} = (x – 2y)^{2}

= (x- 2y)(x – 2y) (b)

**Higher Order Thinking Skills (Hots)**

**Factorise the following**

Question 1.

x^{2} + \(\left(a+\frac{1}{a}\right)\)x + 1

Solution:

x^{2} + \(\left(a+\frac{1}{a}\right)\)x + 1

= x^{2} + ax + \(\frac{x}{a}\) + 1

= x(x + a) + \(\frac{1}{a}\)(x + a)

= (x + a)\(\left(x+\frac{1}{a}\right)\)

Question 2.

36a^{4} – 97a^{2}b^{2} + 36b^{4}

Solution:

= 36a^{4} – 97a^{2}b^{2} + 36b^{4}

= 36a^{4} – 72a^{2}b^{2} + 36b^{4} – 25a^{2}b^{2}

= (6a^{2})^{2} – 2 × 6a^{2} × 6b^{2} + (6b^{2})^{2} – (5ab)^{2}

= (6a^{2} – 6b^{2})^{2} – (5ab)^{2}

= (6a^{2} – 6b^{2} + 5ab)(6a^{2} – 6b^{2} – 5ab)

= (6a^{2} + 5ab – 6b^{2})(6a^{2} – 5ab – 6b^{2})

= [6a^{2} + 9ab – 4ab – 6b^{2}] [6a^{2} – 9ab + 4ab – 6b^{2}]

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

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

Question 3.

2x^{2} – \(\sqrt{3}\)x – 3

Solution:

2x^{2} – \(\sqrt{3}\)x – 3

= 2x^{2} – \(2 \sqrt{3}\)x + \(\sqrt{3}\)x – 3

{∵ 2 × (-3) = -6

∴ -6 = \(-2 \sqrt{3} \times \sqrt{3}\)

–\(\sqrt{3}\) = -2\(\sqrt{3}\) + \(\sqrt{3}\)}

= 2x(x – \(\sqrt{3}\) ) + \(\sqrt{3}\) (x – \(\sqrt{3}\) )

= (x – \(\sqrt{3}\) )(2x + \(\sqrt{3}\))

Question 4.

y(y^{2} – 2y) + 2(2y – y^{2}) – 2 + y

Solution:

y(y^{2} – 2y) + 2(2y – y^{2}) – 2 + y

= y^{3} – 2y^{2} + 4y – 2y^{2} -2 + y

= y^{3} – 4y^{2} + 5y – 2

= y^{3} – 2y^{2} + y – 2y^{2} + 4y – 2

= y(y^{2} -2y + 1) – 2(y^{2} -2y + 1)

= (y^{2} – 2y + 1)(y – 2)

= [(y)^{2} – 2 × y × 1 + (1)^{2}] (y – 2)

= (y – 1)^{2}(y – 2)