Q. 84.1( 9 Votes )

Let us factorise the following algebraic expressions:

3a (3a + 2c) – 4b (b + c)

Answer :

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)


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