ML Aggarwal Factorisation MCQs Class 8 ICSE Ch-11 Maths Solutions

ML Aggarwal Factorisation MCQs Class 8 ICSE Ch-11 Maths Solutions. We Provide Step by Step Answer of  MCQs Questions for Factorisation as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

ML Aggarwal Factorisation MCQs Class 8 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 8th
Chapter-11 Factorisation
Writer ML Aggarwal
Book Name Understanding
Topics Solution of MCQs
Edition 2023-2024

Factorisation MCQs

ML Aggarwal Class 8 ICSE Maths Solutions

Page-200

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 x2 + 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 ax2 + 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.

Answer:

(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 x2 + 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 ax2 + 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 6x2y2 and 8xy3 is 2xy2.
(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 ax2 + 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 4x2 – 12x + 9 are (2x – 3) (2x – 3).

(vii) The value of p for (51) – (49) = 100 p is 2.

(viii) The difference of the squares of two consecutive natural numbers is their sum.

Answer:

(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 6x2y2 and 8xy2 is 2xy2. 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 ax2 + 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
(vi) Factors of 4x2 – 12x + 9 are (2x – 3) (2x – 3). True.

(vii) The value of p for (51) – (49) = 100 p is 2. True.

(viii) The difference of the squares of two consecutive natural numbers is their sum. True.


Factorisation MCQs

ML Aggarwal Factorisation MCQs Class 8 ICSE Maths

Page-200

Choose the correct answer from the given four options (3 to 14):
Question 3. H.C.F. of 6abc, 24ab2, 12a2b is

(a) 6ab
(b) 6ab2
(c) 6a2b
(d) 6abc

Answer:

H.C.F. of babe, 24ab2, 12a2b
= H.C.F. of 6, 24, 12 × H.C.F. of abc, ab2, a2b
= 6 × a × b = 6ab (a)

Question 4. Factors of 12a2b + 15ab2 are

(a) 3a(4ab + 5b2)
(b) 3ab(4a + 5b)
(c) 3b(4a2 + 5ab)
(d) none of these

Answer:

12a2b + 15 ab= 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

Answer:

6xy – 4y + 6 – 9x
= 6xy – 9x – 4y + 6
= 3x(2y – 3) -2(2y – 3)
= (2y – 3) (3x – 2)

Question 6. Factors of 49p3q – 36pq are

(a) p(7p + 6q) (7p – 6q)
(b) q(7p – 6) (7p + 6)
(c) pq(7p + 6) (7p – 6)
(d) none of these

Answer:

49p2q – 36pq
= pq(49p2 – 36)
=pq[(7p)2 – (6)2]
= pq(7p + 6) (7p – 6)


Factorisation MCQs

ML Aggarwal Class 8 ICSE Maths Solutions

Page-201

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

Answer:

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)

Answer:

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

Question 9. Factors of z2 – 4z – 12 are

(a) (z + 6)(z – 2)
(b) (z – 6)(z + 2)
(c) (z – 6)(z – 2)
(d) (z + 6)(z + 2)

Answer:

Factors of z2 – 4z – 12
⇒ z2 – 6z + 2z – 12
= z(z – 6) + 2(z – b)
= (z – 6)(z + 2) (b)

Question 10. Factors of 63a2 – 112b2 are

(a) 63 (a – 2b)(a + 2b)
(b) 7(3a + 2b)(3a – 2b)
(c) 7(3a + 4b)(3a – 4b)
(d) none of these

Answer:

Factors of 63a2 – 112b2 are
= 7(9a2 – 16b2)
= 7[(3a)2 – (4b)2]
= 7(3a + 4b)(3a – 4b) (c)

Question 11. Factors of p4 – 81 are

(a) (p2 – 9)(p2 + 9)
(b) (p + 3)2 (p – 3)2
(c) (p + 3) (p – 3) (p2 + 9)
(d) none of these

Answer:

p4 – 81 = (p2)2 – (9)2
= (p2 + 9)(p2 – 9)
= (p2 + 9){(p)2 – (3)2}
= (p2 + 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)

Answer:

3x2 + 7x – 6
= 3x2 + 9x – 2x – 6
= 3x(x + 3) -2(x + 3)
= (3x – 2) (x + 3) (a)

Question 13. Factors of 16x2 + 40x + 25 are

(a) (4x + 5)(4x + 5)
(b) (4x + 5)(4x – 5)
(c) (4x + 5)(4x + 8)
(d) none of these

Answer:

16x2 + 40x + 25
= (4x)2 + 2 × 4x × 5 + (5)2
= (4x + 5)2
= (4x + 5)(4x + 5) (a)

Question 14. Factors of x2 – 4xy + 4y2 are

(a) (x – 2y)(x + 2y)
(b) (x-2y)(x-2y)
(c) (x + 2y)(x + 2y)
(d) none of these

Answer:

x2 – 4xy + 4y2
= (x)2 – 2 × x × 2y + (2y)2 = (x – 2y)2
= (x- 2y)(x – 2y) (b)


Factorisation (HOTS)

ML Aggarwal Factorisation MCQs Class 8 ICSE Maths

Page-201

Higher Order Thinking Skills (HOTS)Factorise the following

Question 1. x2 + (a + 1/a) x + 1 (a + 1/a)

Answer:

x2 + (a + 1/a) x + 1
= x2 + ax + x/a + 1
= x(x + a) + 1/a(x + a)
= (x + a)(x + 1/a)

Question 2. 36a4 – 97a2b2 + 36b4

Answer:

= 36a4 – 97a2b2 + 36b4
= 36a4 – 72a2b2 + 36b4 – 25a2b2
= (6a2)2 – 2 × 6a2 × 6b2 + (6b2)2 – (5ab)2
= (6a2 – 6b2)2 – (5ab)2
= (6a2 – 6b2 + 5ab)(6a2 – 6b2 – 5ab)
= (6a2 + 5ab – 6b2)(6a2 – 5ab – 6b2)
= [6a2 + 9ab – 4ab – 6b2] [6a2 – 9ab + 4ab – 6b2]
= [3a(2a + 3b) – 2b(2a + 3b)] [3a(2a – 3b) + 2b(2a – 3b)]
= (2a + 3b)(3a – 2b)(2a – 3b)(3a + 2b)

Question 3. 2x2 – √3x – 3

Answer:

2x2 – √3x – 3
= 2x2 – 2√3x + √3x – 3
{∵ 2 × (-3) = -6
∴ -6 = -2√3 x √3

–√3 = -2√3 + √3}
= 2x(x – √3 ) + √3 (x – √3 )
= (x – √3 )(2x + √3)

Question 4. y(y2 – 2y) + 2(2y – y2) – 2 + y

Answer:

y(y2 – 2y) + 2(2y – y2) – 2 + y
= y3 – 2y2 + 4y – 2y2 -2 + y
= y3 – 4y2 + 5y – 2
= y3 – 2y2 + y – 2y2 + 4y – 2
= y(y2 -2y + 1) – 2(y2 -2y + 1)
= (y2 – 2y + 1)(y – 2)
= [(y)2 – 2 × y × 1 + (1)2] (y – 2)
= (y – 1)2(y – 2)

— End of Factorisation MCQs Class 8 ICSE Maths Solutions :–

Return to : ML Aggarwal Maths Solutions for ICSE Class -8

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