## Book: NCERT - Mathematics

### Chapter: 14. Factorisation

#### Subject: Maths - Class 8th

##### Q. No. 2 of Exercises 14.2

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##### Factorise(i) (ii) (iii) (iv) (v) (l + m)2 - (l - m)2(vi) (vii)(viii)

(i) 4p2 - 9q2

4p2 - 9q= (2p)2 - (3q)2

Using identity a2 - b2 = (a + b)(a - b)

Here a = 2p; b = 3q

(2p)2 - (3q)2 = (2p + 3q) (2p – 3q)

(ii) 63a2 - 112b2

7(9a2 - 16b2) = 7{(3a)2 - (4b2)}

Using identity a2 - b2 = (a + b)(a - b)

Here a = 3a; b = 4b

7{(3a)2 - (4b)2} = 7{(3a + 4b) (3a – 4b)}

(iii) 49x2 - 36 = (7x)2 - 62

Using identity a2 - b2 = (a + b)(a - b)

Here a = 7x; b = 6

(7x)2 - (6)2 = (7x + 6) (7x – 6)

(iv) 16x⁵ - 144x3 = 16x3(x2 - 9) 16x3{x2 - 32}

Using identity a2 - b2 = (a + b)(a - b)

Here a = x; b = 3

16x3{(x)2 - (3)2} = 16x3{(x + 3) (x – 3)}

(v) (l + m)2 - (l - m)2

Using identity a2 - b2 = (a + b)(a - b)

Here a = (l + m); b = (l - m)

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

(l + m)2 - (l + m)2 = 2l × 2m = 4lm

(vi) 9x2y2 - 16 = (3xy)2 - 42

Using identity a2 - b2 = (a + b)(a - b)

Here a = 3xy; b = 4

(3xy)2 - 42 = (3xy + 4) + (3xy - 4)

(vii) (x2- 2xy + y2 ) = (x + y)2 [using identity (a + b)2 = a2 + 2ab + b2]

(x + y)2 - z2 =

Using identity a2 - b2 = (a + b)(a - b)

Here a = (x + y)2 ; b = z

(x + y)2 - z2 = (x + y + z) + (x + y - z)

(viii) 25a2 - (4b2 - 28bc + 49c2) (5a)2 - (2b – 7c )2 [using identity (a - b)2 = a2 - 2ab + b2]

Using identity a2 - b2 = (a + b)(a - b)

Here a = 5a ; b = 4b – 7c

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

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