The L.C.M of $2 x + 2, 3 x^{2} - 12$ and $4 x^{2} + 12 x + 8$ is

12(x+1)(x-3)(x+2)

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12(x+1)(x-2)(x+3)

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12(x+1)(x-3)(x+3)

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12(x+1)(x+2)(x-2)

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Solution

The correct option is C 12(x+1)(x+2)(x-2)
We have,
$2 x + 2, 3 x^{2} - 12, 4 x^{2} + 12 x + 8$
$2 (x + 1), 3 (x^{2} - 4), 4 x^{2} + 8 x + 4 x + 8$
$2 (x + 1), 3 (x - 2) (x + 2), 4 x (x + 2) + 4 (x + 2)$
$2 (x + 1), 3 (x - 2) (x + 2), 4 (x + 1) (x + 2)$

$L . C . M$
$= 2 (x + 1) \times 3 (x - 2) (x + 2) \times 4 (x + 1) (x + 2)$
$= 12 (x + 1) (x - 2) (x + 2)$

Hence, this is the answer.

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