How to find value of $k$? I have a question. How can I find value of $k$?
The given below
$$x + 4$$ is a factor of $$x^4+kx^3-4x^2$$
What is  value of $k$?
Thanks for the solutions.
 A: By factor theorem, $(-4)^4+k(-4)^3-4(-4)^2=0$.
Solving, we get
$\begin{align*}256-64k-4(16)&=0\\
64k&=192\\
k&=3\end{align*}$
A: Factor out $x^2$:
$$x^4+kx^3-4x^2=x^2(x^2+kx-4)$$
So, from the given condition ($x+4$ is a factor) one of the two roots of the quadratics in the brackets is $x_1=-4$. 
From the Vieta's:
$$x_1x_2=-4 \Rightarrow -4x_2=-4 \Rightarrow x_2=1;\\
x_1+x_2=-k \Rightarrow -4+1=-k \Rightarrow k=3.$$
A: Two ways. 
1) Hard but straightforward way:
$x^4 + kx^3 -4x^2 = (x+4)(x^3 + ax^2 + bx + c)$ so
$x^4 + kx^3 - 4x^2 = x^4 + (a+4)x^3 + (b+4a)x^2 + (c + 4b)x + 4c$
So $4c = 0; c+4b=0; b+4a=-4; a+4 =k$
So $c = 0$ and $b = 0$ and $a =-1$ and $-1 +4  = 3 = k$.
$x^4 + 3x^3 -4x^2 = (x+4)(x^3 - x^2)$
2) Very easy way
If $x + 4$ is a factor then $-4$ is a root.
So $(-4)^4 + k(-4)^3 -4(-4)^2 = 0$
$4^4 - 4^3k - 4^3 = 0$
Solve for $k$.  $4^3k = 4^4 - 4^3$ 
So $k =\frac {4^4 -4^3}{4^3} =4-1 =3$.
3) Division.
$\frac {x^4 + kx^3 - 4x^2}{x+4} =$
$\frac {x^4 + 4x^3}{x+4} + \frac {(k-4)x^3 -4x^2}{x+2} =$
$x^3 + \frac {(k-4)x^3 + 4(k-4)x^2}{x+4} + \frac {-4(k-4)x^2 -4x^2}{x+2} =$
$x^3 + (k-4)x^2 + \frac {(12-4k)x^2}{x+4} =$
$x^3 + (k-4)x^2 + \frac {(12-4k)x^2+ (48- 16k)x}{x+4}+ \frac {(16k-48)x}{x+4} =$
$x^3 + (k-4)x^2 + (12-4k)x + \frac {(16k-48)x + (64k - 4*48)}{x+4}-\frac {64k-4*48}{x+4} =$
$x^3 + (k-4)x^2 + (12-4k)x + (16k-48)-\frac {64k-4*48}{x+4} =$
But $x+4$ has to divide evenly so $64k - 4*48 = 0$.  Solve for $k$.  $64k = 4*48= 4*16*3 = 64*3$ so $k =3$.
(and we get $(x^3 -x^2)(x+4) = x^4 + 3x^3 - 4x^2$ )
