Find k so that polynomial division has remainder 0 
Find $k$ so that $x^3-kx^2+3x+7k$ has remainder $0$ when divided by $x+2$.

How do I approach this problem? I know how to do polynomial long division and synthetic division, but I don't know how to apply it in this equation.
 A: Hint:
Let $F(x)=x^3-kx^2+3x+7k$. Then $F(-2)=0$
A: Hint:  Question is same as asking when will $x=-2$ be a root of the polynomial. Evaluate your polynomial at $x=-2$ to see for what value of $k$ the expression will vanish. 
A: Divide your polynomial $x^3-kx^2+3x+7k$ by $x+2$ like usual.
When you do, you find that the numerator of the remainder term becomes $7k+2(2k-1)$ (by my quick calculation). Then, we desire that the remainder is $0$, so really the problem is asking us to solve $7k+2(2k-1)=0$ for a suitable k. 
Hope this helps!
A: You know the following:
$$
x^3 - kx^2 + 3x + 7k = (x + 2)(ax^2 + bx + c)
$$
Just do the multiplication and find $a$, $b$, and $c$.
$$
(x + 2)(ax^2 + bx + c) = ax^3 + (2a + b)x^2 + (c + 2b)x + 2c
$$
Now we set each coefficient equal:
$$
a = 1 \\
2a + b = -k \rightarrow b = -k - 2\\
c + 2b = 3 \rightarrow c = 3 - 2(-k - 2) \\
2c = 7k \rightarrow 2(3 - 2(-k - 2) = 7k
$$
Solving the last equation gives:
$$
6 + 4k + 8 = 7k \rightarrow 14 = 3k \rightarrow k = \frac{14}{3}
$$
A: Let's call your polynomial $p(x)$. By the division theorem, you know you $p(x)=(x+2)q(x)+r(x)$, where $q(x)$ is the quotient and $r(x)$ is the remainder term (with degree two or less, why?). 
Then what happens to $p(x)=(x+2)q(x)+r(x)$ when you let $x=-2$? 
A: Since the remainder is $0$, it implies $(x+2)$ is a factor of $x^3-kx^2+3x+7k$  
i.e., $x^3-kx^2+3x+7k = g(x)(x+2)$  
This is valid for all real values of x. Therefore, substituting $x=-2$ in the above equation, we get:
$(-2)^3-k(-2)^2+3(-2)+7k = 0$  
$\implies 3k -14 = 0$  
Hence, $k = \frac{14}{3}$
