Let us factorise the following algebraic expressions:

3a (3a + 2c) – 4b (b + c)

Given, 3a(3a + 2c) – 4b(b + c)

⇒ 9a² + 6ac – 4b² – 4bc

⇒ (3a)² + 2(3)(ac) – (2b)² – 4bc + c² – c²

⇒ (3a)² + 2(3)(ac) + c² – ((2b)² + 2(2b)(c) + c²)

⇒ (3a + c)² – (2b + c)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (3a + c – 2b – c)(3a + c + 2b + c)

⇒ (3a – 2b)(3a + 2b + 2c)