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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.

user376984
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2 Answers2

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$x^3-4x=x(x^2-4)=x(x+2)(x-2)$. 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, $x-2|f(x).$

Soham
  • 9,990
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Yes the degree of $R(x)$ has to be less than the degree of $x-2$.

Degree is a type of Euclidean function (https://en.wikipedia.org/wiki/Euclidean_domain), and the polynomial ring (over the reals) is a Euclidean domain.

So if $a = bq + r$, then $r = 0$ or $deg(r) < deg(b)$, where $r$ is the remainder and $b$ is the divisor.

yoyostein
  • 19,608
  • Then how do I subtract $x^{3}$? In my mind I have to multiply $(x-2)$ with $x^{2}$ in order to get rid of the $x^{3}$. What is left is $2x^{2} - 4x$. The same will happen if you do long division (I think...) – user376984 Oct 27 '16 at 11:42