ICSE

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Check Your Progress

Question 1.
Find the HCF of the given polynomials:
(i) 14pq, 28p²q²
(ii) 8abc, 24ab², 12a²b
Solution:

Question 2.
Factorise the following:
(i) 10x² – 18x³ + 14x⁴
(ii) 5x²y + 10xyz + 15xy²
(iii) p²x² + c²x² – ac² – ap²
(iv) 15(x + y)² – 5x – 5y
(v) (ax + by)² + (ay – bx)²
(vi) ax + by + cx + bx + cy + ay
(vii) 49x² – 70xy + 25y²
(viii) 4a² + 12ab + 9b²
(ix) 49p² – 36q²
(x) 100x³ – 25xy²
(xi) x² – 2xy + y² – z²
(xii) x⁸ – y⁸
(xiii) 12x³ – 14x² – 10x
(xiv) p² – 10p + 21
(xv) 2x² – x – 6
(xvi) 6x² – 5xy – 6y²
(xvii) x² + 2xy – 99y²
Solution:

Question 3.
Divide as directed:
(i) 15(y + 3)(y² – 16) ÷ 5(y² – y – 12)
(ii) (3x³ – 6x² – 24x) ÷ (x – 4) (x + 2)
(iii) (x⁴ – 81) ÷ (x³ + 3x² + 9x + 27)
Solution:

ML Aggarwal Class 8 Solutions for ICSE Maths

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 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 there is ……….. ac.
Solution:

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).
Solution:

Multiple Choice Questions
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
Solution:

Question 4.
Factors of 12a²b + 15ab² are
(a) 3a(4ab + 5b2)
(b) 3ab(4a + 5b)
(c) 3b(4a² + 5ab)
(d) none of these
Solution:

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:

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
Solution:

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:

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:

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)
Solution:

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
Solution:

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
Solution:

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:

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
Solution:

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
Solution:

Higher Order Thinking Skills (Hots)
Factorise the following
Question 1.
x² + \(\left(a+\frac{1}{a}\right)\)x + 1
Solution:

Question 2.
36a⁴ – 97a²b² + 36b⁴
Solution:

Question 3.
2x² – \(\sqrt{3}\)x – 3
Solution:

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

ML Aggarwal Class 8 Solutions for ICSE Maths

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5

Question 1.
Work out the following divisions:
(i) (35x + 28) ÷ (5x + 4)
(ii) 7p²q²(9r – 27) ÷ 63pq(r – 3)
Solution:

Question 2.
Divide as directed:
(i) 6(2x + 7) (5x – 3) ÷ 3(5x – 3)
(ii) 33pq (p + 3) (2q – 5) ÷ 11p (2q – 5)
Solution:

Question 3.
Factorise the expression and divide them as directed:
(i) (7x² – 63x) ÷ 7(x – 3)
(ii) (3p² + 17p + 10) ÷ (p + 5)
(iii) 10xy(14y² + 43y – 21) ÷ 5x(7y – 3)
(iv) 12pqr(6p² – 13pq + 6q²) ÷ 6pq(2p – 3q)
Solution:

ML Aggarwal Class 8 Solutions for ICSE Maths

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4

Factorise the following (1 to 11) polynomials:
Question 1.
(i) x² + 3x + 2,
(ii) z² + 10z + 24
Solution:

Question 2.
(i) y² – 7y + 12
(ii) m² – 23m + 42
Solution:

Question 3.
(i) y² – 5y – 24,
(ii) t² + 23t – 108
Solution:

Question 4.
(i) 3x² + 14x + 8,
(ii) 3y² + 10y + 8
Solution:

Question 5.
(i) 14x² – 23x + 8,
(ii) 12x² – x – 35
Solution:

Question 6.
(i) 6x² + 11x – 10
(ii) 5 – 4x – 12x²
Solution:

Question 7.
(i) 1 – 18y – 63y²,
(ii) 3x² – 5xy – 12y²
Solution:

Question 8.
(i) x² – 3xy – 40y²
(ii) 10p²q² – 21pq + 9
Solution:

Question 9.
(i) 2a²b² + ab – 45
(ii) x (12x + 7) – 10
Solution:

Question 10.
(i) (a + b)² – 11(a + b) – 42
(ii) 8 + 6(p + q) – 5(p + q)
Solution:

Question 11.
(i) (x – 2y)² – 6(x – 2y) + 5
(ii) 7 + 10(2x – 3y) – 8(2x – 3y)²
Solution:

ML Aggarwal Class 8 Solutions for ICSE Maths

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.3

Question 1.
Factorise the following expressions using algebraic identities:
(i) x² – 12x + 36
(ii) 36p² – 60pq + 25q²
(iii) 9y² + 66xy + 121y²
(iv) a⁴ + 6a²b² + 9b⁴
(v) x² + \(\frac{1}{x^{2}}\) + 2
(vi) x² + x + \(\frac{1}{4}\)
Solution:

Factorise the following (2 to 13) expressions:
Question 2.
(i) 4p² – 9
(ii) 4x² – 169y²
Solution:

Question 3.
(i) 9x²y² – 25
(ii) 16x² – \(\frac{1}{144}\)
Solution:

Question 4.
(i) 20x² – 45y²
(ii) \(\frac{9}{16}\) – 25a²b²
Solution:

Question 5.
(i) (2a + 3b)² – 16c²
(ii) 1 – (b – c)²
Solution:

Question 6.
(i) 9 (x + y)² – x²
(ii) (2m + 3n)² – (3m + 2n)²
Solution:

Question 7.
(i) 25 (a + b)² – 16 (a – b)²
(ii) 9 (3x + 2)² – 4 (2x – 1)²
Solution:

Question 8.
(i) x³ – 25x
(ii) 63p²q² – 7
Solution:

Question 9.
(i) 32a²b – 72b³
(ii) 9 (a + b)³ – 25 (a + b)
Solution:

Question 10.
(i) x² – y² – 2y – 1
(ii) p²– 4pq + 4q² – r²
Solution:

Question 11.
(i) 9x² – y² + 4y – 4
(ii) 4a² – 4b² + 4a + 1
Solution:

Question 12.
(i) 625 – p⁴
(ii) 5y⁵ – 405y
Solution:

Question 13.
(i) x⁴ – y⁴ + x² – y²
(ii) 64a² – 9b² + 42bc – 49c²
Solution:

ML Aggarwal Class 8 Solutions for ICSE Maths