Factor completely: $4 x^{3} + 12 x^{2} + 8 x$

x (x + 2) (x + 1))

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4 x (x - 2) (x + 1))

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4 x (x + 2) (x + 1))

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4 x^{2} (x + 2) (x + 1))

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Solution

The correct option is C $4 x (x + 2) (x + 1))$
$4 x^{3} + 12 x^{2} + 8 x$
Take $4 x$ as common,
$= 4 x (x^{2} + 3 x + 2)$
$= 4 x (x^{2} + 2 x + x + 2)$ (replacing $3 x$ into $2 x + x$ to factorise easily)
$= 4 x (x (x + 2) + x (x + 2))$
So, $4 x (x + 2) (x + 1))$ is a complete factor.

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