Question

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

Answer 1

If a polynomial p(x) is divisible by q(x), the roots of q(x) are roots of p(x), too.

In this question, p(x)=x³+6x²+kx+12 and q(x)=x+4. Clearly, the root of q(x) is -4. By what was written initially, -4 is a root of p(x). So:
$p(-4)=0\\\\ (-4)^3+6(-4)^2+k(-4)+12=0\\\\ -64+6\cdot16-4k+12=0\\\\ -64+96-4k+12=0\\\\ 4k=44\\\\ k=\dfrac{44}{4}\\\\ \boxed{k=11}$