# Polynomial Long Division:

My question essentially boils down to this: (it was part of a question about polynomial ring ideals)

Find an integer $$b$$ such that the rational function $$$$\frac{x^5-bx}{x^2-2x}\in \mathbb{Z}[x]$$$$ I've tried plugging in a bunch of values for $$b$$ using a calculator but none of them have come out to whole polynomials. Is there an easier way to do this?

• The numerator must be divisible by $x$ so that $a=0$ must hold. Similarly it must be divisible by $x-2$ too because otherwise it cannot lie in the ring – Ninja Dec 5 '20 at 22:57
• @Ninja yeah I figured, I just can't find a $b$ that gives the result... – Vladimir Lenin Dec 5 '20 at 23:01
• I might edit the question to account for that – Vladimir Lenin Dec 5 '20 at 23:01
• In a polynomial ring $R[X]$, if a polynomial $P(X)$ is divisible by $X-r$ for some $r \in R$ then $P(r)=0$ must hold and in your case $x^4-b$ must be divisible by $x-2$. – Ninja Dec 5 '20 at 23:05

You may be guessing a long time at that rate! Write $$x^5 - bx - a = (x^2-2x)(-------),$$ and fill in the divisor's bracket term by term. To multiply to $$x^5$$, the first term needs to be $$x^3$$: $$x^5 - bx - a = (x^2-2x)(x^3-----).$$ Now $$2x^4$$ is being subtracted, so you better add $$2x^2$$ to the divisor to make that up. $$x^5 - bx - a = (x^2-2x)(x^3+2x^2----).$$ Then $$x^5 - bx - a = (x^2-2x)(x^3+2x^2+4x--).$$ So then $$x^5 - bx - a = (x^2-2x)(x^3+2x^2+4x+8),$$ and you can read off $$b = 16$$, $$a = 0$$.
If $$x^2-2x=x(x-2)$$ divides $$x^5-bx=x(x^4-b)$$, it means $$2$$ is a root of $$x^4-b$$, whence $$b=2^4$$.
Indeed, you easily check that $$x^5-16x=(x^2-2x)(x^3+2x^2+4x+8).$$
Proof $$1\!:\$$ $$x(x\!-\!2)\mid xf \overset{{\rm cancel}\ x}\iff x\!-\!2\mid f \!\!\!\overset{\rm\large\color{#c00}{RT}\!\!\!}\iff f(2)=0 \overset{f = x^4-b\!}\iff b=2^4\,$$ by $$\color{#c00}{\small \rm RT}$$ = Rem Theorem.
Proof $$2\!:\ 0 = xf \bmod x(x\!-\!2) \overset{\color{#90f}{\rm DL}}=\, \color{#0a0}x(f \bmod x\!-\!2) \overset{\rm\large\color{#c00}{RT^{\phantom{|}}}_{\phantom |}\!\!\!}= x f(2)\! \iff\! f(2)=0\!\iff \ldots$$
upon applying $$\color{#90f}{\rm DL}\!: \ gf\bmod gh^{\phantom{|^{|^|}}}\!\!\! =\, \color{#0a0}g(f\bmod h)\,$$ to factor out $$\,\color{#0a0}{g\!=\!x},\,$$ $$\color{#90f}{\rm DL}$$ = mod Distrib. Law.