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Factorisation Using Algebraic Identities

Factorise 4...

Question

Factorise $4 a^{2} + b^{2} + c^{2} - 4 a b + 2 b c - 4 c a$ using identity

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Solution

$4 a^{2} + b^{2} + c^{2} - 4 a b + 2 b c - 4 c a$
= $(2 a)^{2} + (- b)^{2} + (- c)^{2} + 2 (2 a) (- b) + 2 (- b) (- c) + 2 (2 a) (- c)$
We know that $x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 x z = (x + y + z)^{2}$
Here, $x = 2 a$ , $y = - b$ and $z = - c$

So, $4 a^{2} + b^{2} + c^{2} - 4 a b + 2 b c - 4 c a$
$= (2 a - b - c)^{2}$
$= (2 a - b - c) (2 a - b - c)$

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