Remainder of monomial / single-term polynomial division

Question

I need the remainder of the following division. The only thing I managed to get so far is:

\begin{align} & \frac{x^{2017}}{x^2-3x+2}=\frac{x^{2017}+x^{2016}-x^{2016}+\dots+x-x-1}{(x-2)(x-1)} + \frac{1}{(x-2)(x-1)} \\[10pt] = {} & \frac{(x-1)\left(x^{2016}+x^{2015}+\dots+x+1\right)}{(x-2)(x-1)}+\frac{1}{(x-2)(x-1)} \\[10pt] = {} & \frac{(x-1) \left(-2(x^{2016}+x^{2015}+x^{2014}+\dots+1)+3(x^{2016}+x^{2015}+x^{2014}+\dots+1)\right)}{(x-2)(x-1)}+\frac{1}{(x-2)(x-1)} \end{align}

But I still couldn't factor $(x-2)$ if that's even possible.

It seems I should come up with (or prove) a general formula for such recursive polynomials but I the independent $\pm 1$ coefficient at the end for each transition is messing things up.

Or maybe my whole approach is wrong. Euclid's algorithm seems also not so helpful.

Would love a hint.

Answer 1

Euclidean division helps: let $P(x)$ be any polynomial.

• Division by $x-1$ yields a polynomial $Q(x)$ such that $$P(x)=(x-1)Q(x)+P(1).\tag{1}$$
• Division of $Q(x)$ by $x-2$ yield a polynomial $S(x)$ such that $$Q(x)=(x-2)S(x)+Q(2).$$

We deduce that $$P(x)=(x-1)(x-2)S(x)+Q(2)(x-1)+P(1).$$ On the other hand, eq. (1) implies $P(2)=Q(2)+P(1)$, whence $$P(x)=(x-1)(x-2)S(x)+\bigl(P(2)-P(1)\bigr)(x-1)+P(1),$$ so $$P(x)\equiv \bigl(P(2)-P(1)\bigr)(x-1)+P(1)\mod (x-1)(x-2).$$

Answer 2

$x^{2017}=(x^2-3x+2)Q(x)+ax+b$. Now evaluate at $x=1$, then at $x=2$, geting two equations in two unknowns.