<s>[INST] Is this valid? Show that $(x-2)\mid x^{3} - 4x$ Is this valid? Here $\textsf{R}(x)$ stands for the remainder. 

Show that $(x-2)\mid x^{3} - 4x$

I know that I can simply  do this: Let $p(x) = x^{3} - 4x$ then if $(x-2)\mid x^{3} - 4x$ it follows that $p(2) = 0$. So we have $2^{3} - 4(2) = 0$. However I am wondering about the following.
The process (in lack of a better word):
$\textsf{R}(x)_{1} = x^{3} - 4x -x^{2}(x-2) = 2x^{2} - 4x$
$\textsf{R}(x)_{2} = 2x^{2} - 4x - 2x(x-2) = 0$
so
$ x^{3} - 4x = \textsf{R}(x)_{1} + x^{2}(x-2) \Rightarrow 2x^{2} - 4x + x^{2}(x-2) \iff 2x(x-2) +  x^{2}(x-2) = (x-2)(2x + x^{2})$
My question is this: Does $\text{deg} \space \textsf{R}(x)$ have to be less than $\text{deg}(x-2)$ in the process? I get the correct answer.
Any answer is appreciated. 
 [/INST] $x^3-4x=x(x^2-4)=x(x+2)(x-2)$. So, you can clearly see that $x-2$ is a factor.
Or you can do it this way,
Let $f(x)=x^3-4x$. See that $f(2)=0$. Hence, by factor theorem, $x-2|f(x).$ 
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