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:
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

(vi) x2 + x + $$\frac{1}{4}$$

Factorise the following (2 to 13) expressions:
Question 2.
(i) 4p2 – 9
(ii) 4x2 – 169y2
Solution:
(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) 16x2 – $$\frac{1}{144}$$
Solution:
(i) 9x2y2 – 25
= (3xy)2 – (5)2
= (3xy + 5) (3xy – 5)

Question 4.
(i) 20x2 – 45y2
(ii) $$\frac{9}{16}$$ – 25a2b2
Solution:
(i) 20x2 – 45y2
= 5 (4x2 – 9y2)
= 5[(2x)2 – (3y)2]
= 5 (2x + 3y) (2x – 3y)

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

Question 10.
(i) x2 – y2 – 2y – 1
(ii) p2– 4pq + 4q2 – r2
Solution:
(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) 9x2 – y2 + 4y – 4
(ii) 4a2 – 4b2 + 4a + 1
Solution:
(i) 9x2 – y2 + 4y – 4
= 9x2 – (y2 – 4y + 4)
= 9x2 – (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
Solution:
(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
Solution:
(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)