Question

Factorise:

36a² – b² + 20bc – 100c²

Answer 1

Given, 36a² – b² + 20bc – 100c²

36a² – b² + 20bc – 100c² can be written as (6a)² – (b² – 20bc + 100c²)

⇒ (6a)² – {(b)² – 2 × b × 10c + (10c)²}

⇒ (6a)² – (b – 10c)²

Now, applying the identity, 

a² – b² = (a + b) (a – b)

⇒ (6a – b + 10c) (6a + b – 10c)

Hence, the required is (6a – b + 10c) (6a + b – 10c).