If $x^{2} - 4$ is a factor of $2 x^{3} + a x^{2} + b x + 12$ where a and b are constant then the value of a and b are

-3, 8

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3, 8

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-3 -8

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3, -8

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Solution

The correct option is C -3 -8
Given, $x^{2} - 4$ is a factor of $p (x) = 2 x^{3} + a x^{2} + b x + 12$
$\Rightarrow (x - 2) a n d (x + 2)$ are factors of $p (x) = 2 x^{3} + a x^{2} + b x + 12$
Therefore, by factor theorem,
$p (2) = 0, p (- 2) = 0$
$\Rightarrow 16 + 4 a + 2 b + 12 = 0 \Rightarrow 2 a + b = - 14$ ...(i)
$\Rightarrow - 16 + 4 a - 2 b + 12 = 0 \Rightarrow 2 a - b = 2$ ...(ii)
On adding (i) and (ii), we get
$4 a = - 12 \Rightarrow a = - 3$
From (i), $b = - 14 - 2 \times - 3 = - 8$
$∴ a = - 3, b = - 8$
Option C is correct.

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