If $x^{3} + 6 x^{2} + 4 x + k$ is exactly divisible by $x + 2$ , then k is equal to :

-6

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-7

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-8

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-10

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Solution

The correct option is C -8
Divide $x^{3} + 6 x^{2} + 4 x + k$ by $x + 2$

$x + 2$ $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯) x^{3} + 6 x^{2} + 4 x + k ($ $x^{2} + 4 x - 6$
$- (x^{3} + 2 x^{2})$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 4 x^{2} + 4 x + k$
$- (4 x^{2} + 8 x)$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 4 x + k$
$- (- 4 x - 8)$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ k + 8$

As, $x + 2$ is a factor of the given polynomial, so remainder must be $0$ .
So, $k + 8 = 0$ or $k = - 8$
Hence, $C$ is correct.

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