NCERT Class-8 Maths Factorisation Chapter-14.2 Factorise

Question

Factorise.

(i) 4p² – 9q²

(ii) 63a² – 112b²

(iii) 49x² – 36

(iv) 16x⁵ – 144x³

(v) (l + m) ² – (l – m) ²

(vi) 9x² y² – 16

(vii) (x² – 2xy + y²) – z²

(viii) 25a² – 4b² + 28bc – 49c²

Answer 1

(i) 4p² – 9q²

Answer: This can be factorised by using the equation; (a + b)(a – b) = a² – b²
Factors = (2p + 3q)(2p – 3q)

(ii) 63a² – 112b²

Answer: 63a² – 112b² = 7(9a² – 16b²)
= 7(3a + 4b)(3a – 4b)

(iii) 49x² – 36

Answer: (7x + 6)(7x – 6)

(iv) 16x⁵ – 144x³

Answer:16x⁵-144x³
= x³(16x²-144)
= x³(4x+12)(4x-12)

(v) (l + m) ² – (l – m) ²

Answer: (l + m + l – m)(l + m – l + m)
2l x 2m = 4lm

(vi) 9x² y² – 16

Answer: (3xy + 4)(3x – 4)

(vii) (x² – 2xy + y²) – z²

Answer: (x² – 2xy + y²) – z²
= (x – y)² – z²
= (x – y + z)(x – y – z)

(viii) 25a² – 4b² + 28bc – 49c²

Answer: 25a² – 4b² + 28bc – 49c²
= (5a)² – (2b)² + 2 x 2b x 7c – (7c)²
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
This can be further factorised by using (a + b)(a – b) = a² – b²
= (5a + 2b – 7c)(5a – 2b + 7c)