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

Factorisation Class-8 ML Aggarwal ICSE Maths Solutions Chapter-11. We provide step by step Solutions of Exercise / lesson-11 Factorisation Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with Exe-11.1 , Exe-11.2, Exe-11.3 Exe-11.4, Exe-11.5, Objective Type Questions (including Mental Maths Multiple Choice Questions ), HOT and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

Factorisation Class-8 ML Aggarwal ICSE Maths Solutions Chapter-11


-: Select Topics :-

Exercise 11.1 ,

Exercise-11.2,

Exercise-11.3,

Exercise-11.4,

Exercise-11.5,

Objective Type Questions,

Mental Maths,

Multiple Choice Questions ,(MCQ)

HOTS

Check Your Progress

Factorisation Ex 11.1 Class-8 ML Aggarwal ICSE Maths Solutions

Factories the following (1 to 8) polynomials:
Question 1.
(i) 8xy3 + 12x2y2
(ii) 15ax3 – 9ax2

Answer-1

(i) 8xy3 + 12x2y2 = 4xy2 (2y + 3x)
(ii) 15ax3 – 9ax2 = 3ax2 (5x – 3)

Question -2.
(i) 21 py2 – 56py
(ii) 4x3 – 6x2

Answer-2

(i) 21 py2 – 56py = 7py (3y – 8)
(ii) 4x3 – 6x2 = 2x2 (2x – 3)

Question 3.
(i) 25abc2 – 15a2b2c
(ii) x2yz + xy2z + xyz2

Answer-3

(i) 25abc2 – 15a2b2c = 5abc (5c – 3ab)
(ii) x2yz + xy2z + xyz2 = xyz(x + y + z)

Question -4.
(i) 8x3 – 6x2 + 10x
(ii) 14mn + 22m – 62p

Answer-4

(i) 8x3 – 6x2 + 10x = 2x (4x2 – 3x + 5)
(ii) 14mn + 22m – 62p = 2 (7mn + 11m – 31p)

Question -5.
(i) 18p2q2 – 24pq2 + 30p2q
(ii) 27a3b3 – 18a2b3 + 75a3b2

Answer-5

(i) 18p2q2 – 24 pq2 + 30p2q
= 6pq (3pq -4q + 5p)
(ii) 27a3b3 – 18a2b3 + 75a3b2
= 3a2b2 (9ab – 6b + 25a)

Question -6.
(i) 15a (2p – 3p) – 106 (2p – 3q)
(ii) 3a (x2 + y2) + 6b (x2 + y2)

Answer-6

(i) 15a (2p – 3q) – 10b (2p – 3q)
= (2p – 3q)(15a – 10b)
= (2p – 3q) (5) (3a – 2b)
= 5 (2p- 3q) (3a – 2b)

(ii) 3a (x2 + y2) + 66 (x2 + y2)
= (x2 + y2) (3a + 6b)
= (x2 + y2) (3) (a + 2b)
= 3 (x2 + y2) (a + 2b)

Question -7.
(i) 6(x + 2y)3 + 8(x + 2y)2
(ii) 14(a – 3b)3 – 21p(a – 3b)

Answer-7

(i) 6(x + 2y)3 + 8(x + 2y)2
(x + 2y)2 [6 (x + 2y) + 8] = (x + 2y)2 [6x + 12y + 8] = (x + 2y)2 (2) (3x + 6y + 4)
= 2 (x + 2y)2 (3x + 6y + 4)

(ii) 14(a – 3b)3 – 21 p(a – 3b)
= 7 [2 (a – 3b)3 -3p(a- 3b)] = 7 [(a – 3b) {2 (a – 3b)2 – 3p}] = 7 (a – 3b) [2 (a – 3b)2 – 3p]

Question -8.
10a (2p + q)3 – 15b (2p + q)2 + 35(2p + q)

Answer-8

10a (2p + q)3 – 15b (2p + q)2 + 35(2p + q)
= 5 [2a (2p + q)]3 – 3b (2p + q)2 + 7 (2p + q)
= 5(2p + q) [2a (2p + q)2 – 3b(2p + q) + 7]


Factorisation Class-8 ML Aggarwal ICSE Maths Solutions Ex 11.2

 Factorise the following (1 to 11) polynomials:

Question -1.
(i) x2 + xy – x – y
(ii) y2 – yz – 5y + 5z

Answer-2

(i) x2 + xy – x – y
= x (x + y) -1 (x + y) = (x + y)(x – 1)
(ii) y2 – yz – 5y + 5z
= y(y – z) -5(y – z)
= (y – z)(y – 5)

Question -2.
(i) 5xy + 7y – 5y2 – 7x
(ii) 5p2 – 8pq – 10p + 16q

Answer-3

(i) 5xy + 7y – 5y2 – 7x
= 5xy – 5y2 + 7y – 7x
= 5y(x – y) -7 (x – y)
= (x – y)(5y – 1)

(ii) 5p2 – 8pq – 10p + 16q
= 5p2 – 10p – 8pq + 16q
= 5p (p – 2) – 8q (p – 2)
= (p – 2) (5p – 5q)
= (5p – 8q)(p – 2)

Question -3.
(i) a2b – ab2 + 3a – 3b
(ii) x3 – 3x2 + x – 3

Answer-3

(i) a2b – ab2 + 3a – 3b
= ab (a – b) + 3 (a – b) = (a – b) (ab + 3)
(ii) x3 – 3x2 + x – 3
= x2 (x – 3) + 1 (x – 3)
= (x – 3) (x2 + 1)

Question -4.
(i) 6xy2 – 3xy – 10y + 5
(ii) 3ax – 6ay – 8by + 4bx

Answer-4

(i) 6xy2 – 3xy – 10y + 5
3xy(2y – 1) -5(2y – 1)
= (2y – 1) (3xy – 5)
(ii) 3ax – 6ay – 8by + 4bx
= 3ax – 6ay + 4bx – 8by
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)

Question- 5.
(i) x2 + xy (1 + y) + y3
(ii) y2 – xy (1 – x) – x3

Answer-5

(i) x2 + xy (1 + y) + y3
= x2 + xy + xy2 + y3
= x(x + y) + y2(x + y)
= (x + y) (x + y2)

(ii) y2 – xy (1 – x) – x3
= y2 – xy + x2y – x3
= y(y – x) + x2 (y – x)
= (y – x) (y + x2)

Question -6.
(i) ab2 + (a – 1) b – 1
(ii) 2a – 4b – xa + 2bx

Answer-6

(i) ab2 + (a – 1) b – 1
= ab2 + ab – b – 1
= ab (b + 1) -1 (b + 1)
= (b + 1) (ab – 1)

(ii) 2a – 4b – xa + 2bx
= 2 (a – 2b) -x (a – 2b)
= (a – 2b) (2 – x)

Question- 7.
(i) 5ph – 10qk + 2rph – 4qrk
(ii) x2 – x(a + 2b) + 2a2

Answer-7

(i) 5ph – 10qk + 2rph – 4qrk
= 5 (ph – 2qk) + 2r (ph – 2qk)
= (ph – 2qk) (5 + 2r)

(ii) x2 – x(a + 2b) + 2ab
= x2 – xa – 2bx + 2ab
= x(x – a) – 2b(x – a)
= (x – a) (x – 2b)

Question -8.
(i) ab (x2 + y2) – xy (a2 + b2)
(ii) (ax + by)2 + (bx – ay)2

Answer-8

(i) ab (x2 + y2) – xy (a2 + b2)
= abx2 + aby2 – a2xy – b2xy
= (abx2 – b2xy) + (aby2 – a2xy)
= bx (ax – by) – ay (ax – by)
= (ax – by) (bx – ay)

(ii) (ax + by)2 + (bx – ay)2
= (a2x2 + b2y2 + 2abxy) + (b2x2 + a2y2 – 2abxy)
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + a2y2 + b2x2 + a2y
= a2 (x2 + y2) + b2 (x2 + y2)
= (a2 + b2) (x2 + y2)

Question 9.
(i) a3 + ab(1 – 2a) – 2b2
(ii) 3x2y – 3xy + 12x – 12

Answer-9

(i) a3 + ab – 2a2b – 2b2
= a3 + ab – 2a2b – 2b2
= a (a2 + b) – 2b (a2 + b)
= (a2 + b) (a – 2b)

(ii) 3x2y – 3xy + 12x- 12
= 3 (x2y – xy + 4x – 4)
= 3 [xy (x – 1) +4 (x – 1)] = 3 (x – 1) (xy + 4)

Question- 10.
(i) a2b + ab2 – abc – b2c + axy + bxy
(ii) ax2 – bx2 + ay2 – by2 + az2 – bz2

Answer-10

(i) a2b + ab2 – abc – b2c + axy + bxy
= ab (a + b) – bc (a + b) + xy (a + b)
= (a + b) (ab – bc + xy)

(ii) ax2 – bx2 + ay2 – by2 + az2 – bz2
= x2 (a – b) + y2 (a – b) + z2 (a – b)
= (a – b)(x2 + y2 + z2)

Question -11.
(i) x – 1 – (x – 1)2 + ax – a
(ii) ax + a2x + aby + by – (ax + by)2

Answer-11

(i) x – 1 – (x – 1)2 + ax – a
= (x – 1) – (x – 1)2 + a (x – 1)
= (x – 1) [1 – (x – 1) + a] = (x – 1) (1 – x + 1 + a)
= (x- 1) (2 – x + a)

(ii) ax + a2x + aby + by – (ax + by)2
= (ax + by) + (a2x + aby) – (ax + by)2
= (ax + by) + a (ax + by) – (ax + by)2
= (ax + by) [1 + a – (ax + by)] = (ax + by) (1 + a – ax – by)


ML Aggarwal Solutions for ICSE Class-8 APC Understanding Maths Factorisation Ex 11.3

Question- 1.
Factories 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}

Answer-1

Using (a + b)2 = a2 + 2ab +b2 and (a – b)2 = a2 – 2ab + b2
(i) y2 – 12x + 36
= (x)2 – 2 × x × 6 + (6)22
= (x – 6)2

(ii) 36p2 – 60pq + 25q2
= (6p)2 – 2 × 6p × 5q + (5q)2
= (6p – 5q)2

(iii) 9x2 + 66xy + 121 y2
= (3x)2 + 2 × 3x × 11y + (11y)2
= (3x + 11 y)2

(iv) a4 + 6a2b2 + 9b4
= (a2)2 + 2 × 2a2 × 3b2 + (3b2)2
= (a2 + 3b2)2

(v) x2 + \frac{1}{x^{2}} + 2
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.3 1

(vi) x2 + x + \frac{1}{4}

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.3 2

Factorise the following (2 to 13) expressions:
Question- 2.

(i) 4p2 – 9
(ii) 4x2 – 169y2

Answer-2

(i) 4p2 – 9
= (2p)2 – (3)2
= (2p + 3) (2p – 3)

(ii) 4x2 – 169y2
= (2x)2 – (13y)2
= (2x + 13y) (2x – 13y)

Question -3.
(i) 9x2y2 – 25
(ii) 16×2 – \frac{1}{144}

Answer-3

(i) 9x2y2 – 25
= (3xy)2 – (5)2
= (3xy + 5) (3xy – 5)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.3 3

Question -4.
(i) 20×2 – 45y2
(ii) \frac{9}{16} – 25a2b2

Answer-4


(i) 20×2 – 45y2
= 5 (4×2 – 9y2)
= 5[(2x)2 – (3y)2] = 5 (2x + 3y) (2x – 3y)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.3 4

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

Answer-5

(i) (2a + 3b)2 – 16c2
= (2a + 3b)2 – (4c)2
= (2a + 3b + 4c) (2a + 3b – 4c)

(ii) 1 – (b – c)2
= (1)2 – (b – c)2
= [1 + b – c)] [1 – (b – c)] = (1 +b – c)(1 – b + c)

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

Answer-6

(i) 9 (x + x)2 – x2
= [3 (x + y)]2 – [x]2
= [3 (x + y) + x] [3 (x + y) – x] = (3x + 3y + x) (3x + 3y – x)
= (4x + 3y) (2x + 3x)

(ii) (2m + 3n)2 – (3m + 2n)2
= (4m2 + 9n2 + 12mn) – (9m2 + 4n2 + 12mn)
= 4m2 + 9n2 + 12mn – 9m2 – 4m2 – 12mn
= 4m2 + 9n2 – 9m2 – 4n2
= – 5m2 + 5n2 = 5 (n2 – m2)
= 5 (m + n) (n – m)

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

Answer-7

(i) 25 (a + b)2 – 16 (a – b)2
= [5 (a + b)]2 – [4 (a – b)]2
= (5a + 5b)2 – (4a – 4b)2
= [(5a + 5b)2 + (4a – 4b)] [(5a + 5b) – (4a – 4b)] = (5a + 5b + 4a – 4b) (5a + 5b – 4a + 4b)
= (9a + ft) (a + 9ft)

(ii) 9 (3x + 2)2 – 4 (2x – 1)2
= [3 (3x + 2)]2 – [2 (2x – 1)]2
= (9x + 6)2 – (4x – 2)2
= [(9x + 6) + (4x – 2)] [(9x + 6) – (4x – 2)] = (9x + 6 + 4x – 2) (9x + 6 – 4x + 2)
= (13x + 4) (5x + 8)

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

Answer-8

(i) x3 – 25x
= x (x2 – 25) = x [(x)2 – (5)2] = x (x + 5) (x – 5)

(ii) 63p2q2 – 7
= 7 (9p2q2 – 1)
= 7 [(3pq)2 – (1)2] = 7 (3pq + 1) (3pq – 1)

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

Answer-9

(i) 32 a2b – 72b3
= 8b (4a2 – 9b2) ⇒ 8b [(2a)2 – (3b)2] = 8b (2a + 3b) (2a – 3b)

(ii) 9 (a + b)3 – 25 (a + b)
= (a + b) [9 (a + b)2 – 25] = (a + b) [{3 (a + b)}2 – (5)2] = (a + 6) [(3a + 3b)2 – (5)2] = (a + b) [(3a + 3b + 5) (3a + 36 – 5)] = (a + b) (3a + 3b + 5) (3a + 3b – 5)

(i) x2 – y2 – 2y – 1
(ii) p2– 4pq + 4q2 – r2

Answer-10

(i) x2 – y2 – 2y – 1
= x2 – (y2 + 2y + 1)
= (x)2 – (y + 1)2
= [x + (y + 1)] [x – (y + 1)] = (x + y + 1)(x – y – 1)

(ii) p2 – 4pq + 4q2 – r2
= (p)2 – 2 × p × 2q + (2q)2 – r2
{∵ (a – b)2 = a2 – 2ab + b2
a2 – b2 = (a + b)(a – b)}
= (p – 2q)2 – (r)2
= (p – 2q + r)(p – 2q – r)

Question 11.
(i) 9×2 – y2 + 4y – 4
(ii) 4a2 – 4b2 + 4a + 1

Answer

(i) 9×2 – y2 + 4y – 4
= 9×2 – (y2 – 4y + 4)
= 9×2 – (y – 2)2
= (3x)2 (y – 2)2
= {3x + (y – 2)} {3x – (y – 2)}
= (3x + y – 2) (3x – y + 2)

(ii) 4a2 – 4b2 + 4a + 1
= (4a2 + 4a + 1) – 4b2
= (2a + 1)2 – (2b)2
= (2a + 2b + 1) (2a – 2b + 1)

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

Answer

(i) 625 – p4
= (25)2 – (p2)2
= (25 + p2) (25 – p2)
= (25 + p2) [(5)2 – (p)2] = (25 +p2) (5 + p) (5 – p)

(ii) 5y5 – 405y
= 5y(y4 – 81)
= 5y [(y2)2 – (9)2] = 5y (y2 + 9) (y2 – 9)
= 5y (y2 + 9) [(y)2 – (3)2
= 5y (y2 + 9) (y + 3) (y – 3)

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

Answer-13

(i) x4 – y4 + x2 – y2
= [(x2)2 – (y2)2] + (x2 – y2)
{a2 – b2 = (a + b) (a – b)}
= (x2 + y2) (x2 – y2) + 1(x2 – y2)
= (x2 – y2) (x2 + y2 + 1)
= (x + y(x – y)(x2 + y2 + 1)

(ii) 64a2 – 9b2 + 42bc – 49c2
= 64a2 – [9b2 – 42bc + 49c2] = (8a)2 – [(3b)2 – 2 × 3b × 7c + (7c)2] {∵ a2 + b2 – 2ab = (a – b)2

a2 – b2 = (a + b)(a – 2)}
= (8a)2 – (3b – 7c)2
= (8a + 3b – 7c) (8a – 3b + 7c)


Ex 11.4 Factorisation ML Aggarwal Class 8 Solutions for ICSE Maths

Factorise the following (1 to 11) polynomials:

Question 1.
(i) x2 + 3x + 2,
(ii) z2 + 10z + 24

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q1
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q1.1

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

Answer

Maths Questions for Class 8 ICSE With Answers Chapter 11 Factorisation Ex 11.4 Q2

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q3

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q4

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q5

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q6

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

Answer

Maths Questions for Class 8 ICSE With Answers Chapter 11 Factorisation Ex 11.4 Q7

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q8

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q9

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

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.4 Q10

(ii) 8 + 6(p + q) – 5(p + q)2
Let p + q = x, then
8 + 6x – 5x2 = -5x2 + 6x + 8
= -(5x2 – 6x – 8)
= [5x2 – 10x + 4x – 8] {∵ 5 × (-8) = 40
∴ -40 = -10 × 4
-6 = -10 + 4}
= (x – 2) (5x + 4)
Substituting the value of x, then
= -(p + q – 2) (5p + 5q +4)
= (4 + 5p + 5q) (-p – q + 2)
= (4 + 5p + 5 q) (2 – p – q)

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

Answer

(i) (x – 2y)2 – 6 (x – 2y) + 5
Let x – 2y = z
Then, (x – 2y)2 – 6 (x – 2y) + 5
= z2 – 6z + 5
∴ z2 – 6z + 5 = z2 – 5z – z + 5
= z(z – 5) – 1 (z – 5)
= (z – 5)(z – 1)
Substituting z = x – 2y, we get,
= [(x-2y) – 5] [(x – 2y) – 1] = (x – 2y – 5) (x – 2y – 1)

(ii) 7 + 10 (2x – 3y) – 8 (2x – 3y)2
Let 2x – 3y = z
Then, 7 + 10 (2x – 3y) – 8 (2x – 3y)2
= 7 + 10z – 8z2
∴ 7 + 10z – 8z2 = 7 + 14z – 4z – 8z2
= 7 (1 + 2z) – 4z (1 + 2z)
= (1 + 2z) (7 – 4z)
Substituting z = 2x – 3y, we get,
= [(1 + 2 (2x – 3y)] [7 – 4 (2x – 3y)] = (1 + 4x – 6y) (7 – 8x + 12y)


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)

Answer

(i) (35x + 28) ÷ (5x + 4)
\frac{7(5 x+4)}{(5 x+4)}=7
(ii) 7p2q2(9r – 27) ÷ 63pq(r – 3)
\frac{7 p^{2} q^{2} \times 9(r-3)}{63 p q(r-3)}
= p2-1 q2-1 × 9 = 9pq

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

Answer

(i) 6(2x + 7) (5x – 3) ÷ 3(5x – 3)
\frac{6(2 x+7)(5 x-3)}{3(5 x-3)} = 2(2x + 7)
(ii) 33pq (p + 3) (2q – 5) ÷ 11p (2q – 5)
\frac{33 p q(p+3)(2 q-5)}{11 p(2 q-5)} = 3q(p + 3)

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)

Answer

(i) (7x2 – 63x) ÷ 7(x – 3)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5 1
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5 2
(ii) (3p2 + 17p + 10 ) ÷ (p + 5)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5 3

(iii) 10xy(14y2 + 43y – 21) ÷ 5x(7y – 3)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5 4
(iv) 12pqr(6p2 – 13pq + 6q2) ÷ 6pq(2p – 3q)
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5 5
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.5 6


Objective Type Questions ML Aggarwal Class 8 Chapter 11 Factorisation

 

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).

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
Correct :
A + B should be equal to ft and AB = ac
(vi) Factors of
4x2 – 12x + 9 are (2x – 3) (2x – 3). True


Multiple Choice Questions (MCQs)

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)

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)


Higher Order Thinking Skills (HOTS)


Factorise the following
Question 1.
x2 + \left(a+\frac{1}{a}\right)x + 1

Answer

x2 + \left(a+\frac{1}{a}\right)x + 1
= x2 + ax + \frac{x}{a} + 1
= x(x + a) + \frac{1}{a}(x + a)
= (x + a)\left(x+\frac{1}{a}\right)

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 – \sqrt{3}x – 3

Answer

2x2 – \sqrt{3}x – 3
= 2x2 – 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(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)


 Chapter 11 Factorisation Check Your Progress

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

Answer

(i) 14pq, 28p2q2 (HCF of 14, 28 = 14)
HCF of 1 4pq, 28p2q2 = 14pq
(ii) 8abc, 24ab2, 12a2b
HCF of 8, 24, 12 = 4
HCF of 8abc, 24ab2, 12a2b = 4ab

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

Answer

(i) 10x2 – 18x3 + 14x4
HCF of 10, 18, 14 = 2
∴ 10x2 – 18x3/sup> + 14 x 4
= 2×2 (5 – 9x + 7×2)

(ii) 5x2y + 10xyz + 15xy2
HCF of 5, 10, 15 = 5
∴ 5x2y + 10xyz + 15xy2
= 5xy(x + 2z + 3y)

(iii) p2x2 + c2x2 – ac2 – ap
= p2x2 – ap2 + c2x2 – ac2
p2(x2 – a) + c2(x2 – a)
= (x2 – a) (p2 + c2)

(iv) 15(x + y)2 – 5x – 5y
= 15(x + y)2 – 5(x + y)
= 5(x + y) [3(x + y) – 1] = 5(x + y) (3x + 3y – 1)

(v) (ax + by)2 + (ay – bx)2
a2x2 + b2y2 + 2abxy + a2y2 + b2x2 – 2abxy
= a2x2 + a2y2 + b2x2 + b2y2
= a2(x2 + y2) + b2(x2 + y2)
= (x2 + y2) (a2 + b2)

(vi) ax + by + cx + bx + cy + ay
= ax + bx + cx + ay + by + cy (grouping)
= x(a + b + c) + y(a + b + c)
= (a + b + c) (x + y)

(vii) 49x2 – 70xy + 25y2
= (7x)2 – 2 × 7x × 5y + (5y)2
{∵ (a – b)2 = a2 – 2ab + b2}
= (7x – 5y)2

(viii) 4a2 + 12ab + 9b2
= (2a)22 + 2 × 2a × 3b + (3b)2
{∵ (a + b)2 = a2 + 2ab + b2}
= (2a + 3b)2

(ix) 49p2 – 36q2
= (7p)2 – (6q)2
= (7p + 6q) (7p – 6q)
{∵ a2 – b2 = (a + b) (a – b)}

(x) 100×3 – 25xy2
= 25x(x2 – y2) = 25x{(x)2 – (y)2}
= 25x(x + y) (x – y)

(xi) x2 – 2xy + y2 – z2
= (x – y)2 – (z)2
{∵ a2 -2ab + b2 = (a – b)2)
a2 – b2 = (a + b)(a – b)}
= (x – y + z)(x – y – z)

(xii) x8 – y8
= (x4)2 – (y4)2
{∵ a2 – b2 = (a + b)(a- b)}
= (x4 + y4)(x4 – y4)
= (x4 + y4) {(x2)2 – (y2)1}
= (x4 + y4) (x2 + y2) (x2 – y2)
= (x4 + y4 (x2 + y2) (x + y) (x – y)

(xiii) 12×3 – 14×2 – 10x
= 2x(6×2 – 7x – 5)
{∵ 6 × (-5) = -30
∴ -30 = -10 × 3
-7 = -10 + 3}
= 2x{6×2 + 3x – 10x – 5}
= 2x{3x(2x + 1) – 5(2x + 1)}
= 2x(2x + 1) (3x – 5)

(xiv) p2 – 10p + 21
= p2 – 3p – 7p + 21
{∵ 21 =-3 × (-7)
-10 = -3 – 7}
= p(p – 3) – 7(p – 3)
= (p – 3)(p – 7)

(xv) 2×2 – x – 6
= 2×2 – 4x + 3x – 6
{ ∵ -6 × 2 = -12
∴ -12 = -4 × 3
-1 = -4 + 3}
= 2x(x – 2) + 3(x – 2)
= (x – 2) (2x + 3)

(xvi) 6×2 – 5xy – 6y2
= 6×2 – 9xy + 4xy – 6y2
{∵ 6 × (-6) = -36
∴ – 36 = -9 × 4
– 5 = -9 + 4}
= 3x(2x – 3y) + 2y(2x – 3y)
= (2x – 3y) (3x + 2y)

(xvii) x2 + 2xy – 99y2
= x2 + 11xy – 9xy – 99y2
{∵ -99 = -11 × 9
-2 = -11 + 9 }
= x(x + 11y) – 9y(x + 11y)
= (x + 11y) (x – 9y)

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

Answer


(i) 15(y + 3)(y2 – 16) ÷ 5(y2 – y – 12)
y2 – 16 = (y)2 – (4)2
= (y + 4)(y – 4)
y2 – y – 12 = y2 – 4y + 3y – 12
= y(y – 4) + 3(y – 4)
= (y – 4)(y + 3)
Now, \frac{15(y+3)\left(y^{2}-16\right)}{5\left(y^{2}-y-12\right)}
\frac{15 \times(y+3)(y+4)(y-4)}{5(y-4)(y+3)}
= 3(y + 4)

(ii) (3×3 – 6×2 – 24x) ÷ (x – 4) (x + 2)
3×3 – 6×2 – 24x = 3x(x2 – 2x – 8)
= 3x{x2 – 4x + 2x – 8}
= 3x{x(x – 4) + 2(x – 4)}
= 3x(x – 4) (x + 2)
∴ \frac{3 x^{3}-6 x^{2}-24 x}{(x-4)(x+2)}=\frac{3 x(x-4)(x+2)}{(x-4)(x+2)}=3 x

(iii) (x4 – 81) ÷ (x3 + 3×2 + 9x + 27)
x4 – 81 = (x2)2 – (9)2 = (x2 + 9) (x2 – 9)
= (x2 + 9) {(x)2 – (3)2}
= (x2 + 9) (x + 3) (x – 3)
x3 + 3×2 + 9x + 27
= (x)2 + (x + 3) + 9 (x + 3)
= (x2 + 9) (x + 3)
Now, \frac{x^{4}-81}{x^{3}+3 x^{2}+9 x+27}
\frac{\left(x^{2}+9\right)(x+3)(x-3)}{\left(x^{2}+9\right)(x+3)}
= x = -3

— End of Factorisation Class-8 ML Aggarwal Solutions :–

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

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