If $x^{3} + 6 x^{2} + 4 x + k$ is exactly divisible by $(x + 2)$ , then the value of k is -

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Solution

The correct option is D
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Since $(x + 2)$ is a factor, so $x = - 2$ will make it zero.

∴ ${(- 2)}^{3} + 6 {(- 2)}^{2} + 2 (- 2) + k = 0$

or k = -12

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