# Find k if (x^3+6x^2+kx+12) is divisible by ((x+4))?

Apr 1, 2017

THe value of $k = 11$

#### Explanation:

We apply the remainder theorem

Let $f \left(x\right) = {x}^{3} + 6 {x}^{2} + k x + 12$

If $f \left(x\right)$ is divisible by $\left(x + 4\right)$, then

$f \left(- 4\right) = 0$

$f \left(- 4\right) = - 64 + 96 - 4 k + 12 = 0$

$4 k = 108 - 64 = 44$

$k = \frac{44}{4} = 11$