Q9 of 65 Page 137

Let us factorise the following algebraic expressions:

a2 – 6ab + 12bc – 4c2

Given, a2 – 6ab + 12bc – 4c2


⇒ a2 – 4c2 – 6ab + 12bc


⇒ (a2 – (2c)2) – 6b(a – 2c)


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


⇒ (a + 2c)(a – 2c) – 6b(a – 2c)


⇒ (a – 2c)(a + 2c – 6b)


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