Question

Factorise:

9b² – 4a² + 20a – 25

Answer 1

Given: 9b² – 4a² + 20a – 25

9b² – 4a² + 20a – 25 can be written as (3b)² – (4a² – 20a + 25)

⇒ (3b)² – {(2a)² – 2 × 2a × 5 + (5)²}

⇒ (3b)² – (2a – 5)²

Now, applying the identity, 

a² – b² = (a + b) (a – b)

⇒ (3b – 2a + 5) (3b + 2a – 5)

Hence, the required is (3b – 2a + 5) (3b + 2a – 5).