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

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Solution

Since (x+2) is a factor, so x = -2 will make it zero.

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

or k = -8

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