Factorise.

 (i) 4p ² – 9q ²                           (ii) 63a ² – 112b ² 

(iii) 49x ² – 36                           (iv) 16x ⁵ – 144x ³ 

(v) (l + m) ² – (l – m) ²             (vi) 9x ² y ² – 16 

(vii) (x ² – 2xy + y ² ) – z²       (viii) 25a ² – 4b ² + 28bc – 49c ²

In this question, we factorise the each of the expression by using the following identities:

(a + b) (a – b) = a ² – b ²

(a – b) ² = a ² – 2ab + b ² 

(i) 4p ² – 9q ²   :

By using the identity, (a + b) (a – b) = a ² – b ²

We get, 

4p ² – 9q ²= (2p $+$9q) (2p $-$ 3q) 

= (2p) ² – (3q) ²

(ii) 63a ² – 112b²:

By using the identity, (a + b) (a – b) = a ² – b ²

We get, 

63a ² – 112b²= 7[(9a) ² – (16b)²]

= 7[3(a) ² – 4(b)²]

= 7 [(3a $+$4b) (3a $-$ 4b)]

(iii) 49x ² – 36  :

By using the identity, (a + b) (a – b) = a ² – b ²

We get, 

49x ² – 36= (7x) ² – (6) ²

= (7x $+$6) (7x $-$ 6) 

 (iv) 16x ⁵ – 144x ³ :

By using the identity, (a + b) (a – b) = a ² – b ²

We get, 

16x ⁵ – 144x ³= (4x³) ² – (12x) ²

= ( 4x³$+$12x) (4x³ $-$ 12x) 

(v) (l + m) ² – (l – m) ²  :

By using the identity, (a + b) (a – b) = a ² – b ²

We get, 

(l + m) ² – (l – m) ²=  (l $+$  m $-$ l $+$ m)(l $+$ m $+$ l $-$ m)

= 2m $\times$ 2I

= 4Im

(vi) 9x ² y ² – 16 :

By using the identity, (a + b) (a – b) = a ² – b ²

We get, 

 9x ² y ² – 16= (3xy) ² – (4) ²

= ( 3xy$+$4) (3xy $-$ 4) 

(vii) (x ² – 2xy + y ² ) – z²  :

By using the identity, (a + b) (a – b) = a ² – b ²

and (a – b) ² = a ² – 2ab + b ²

We get, 

(x ² – 2xy + y ² ) – z²= (x  $-$ y)² – (z) ²

= ( x$-$y $+$ z) (x $-$ y $-$ z) 

(viii) 25a ² – 4b ² + 28bc – 49c ²:

By using the identity, (a + b) (a – b) = a ² – b ²

and (a – b) ² = a ² – 2ab + b ²

We get, 

25a ² – 4b ² + 28bc – 49c ²= 25a ²$-$(4b ² + 28bc – 49c ²)

=( 5a)² $-$ [(2b)²$-$ 2  $\times$ 2b $\times$ 7c $+$ (7c)²]

= (5a)² $-$ [2b$-$ 7c]² 

 = ( 5a$+$(2b$-$ 7c)) (5a$-$(2b$-$ 7c) ) 

= ( 5a$+$2b$-$ 7c)) (5a$-$2b$+$ 7c))