Question

Factorize :
(i) 6ab – b² + 12ac – 2bc
(ii) 9(2a – b)² – 4(2a – b) – 13​

Answer 1

Answer:

Answer

Given expression is 6ab−b2 + 2+12ac−2bc

It can be written as 6ab−b

2 +12ac−2bc=b(6a−b)+2c(6a−b)=(b+2c)(6a−b)

Thus factors are,(b+2c),(6a−b).

Answer 2

Answer::-

Factorize :

Factorize :(i) 6ab – b² + 12ac – 2bc

Given expression is 6ab−b²+12ac−2bc

It can be written as 6ab−b²+12ac−2bc

=b(6a−b+2c(6a−b)=(b+2c)(6a−b)

Thus factors are,(b+2c),(6a−b).

ii) 9(2a – b)² – 4(2a – b) – 13

9(2a−b)²−4(2a−b)−13

Then,=((2a−b)+1)(9(2a−b)−13)

∴9(2a−b)²−4(2a−b)−13=(2a−b+1)(18a−9b−13)