Dividing polynomials such that we get the desired remainder?

Question

Find $a,b,c$ such that the remainder of $$\frac{a(x^{11}-x^4+1)^{15}-x^2+bx-c}{x^3+x}$$ is equal to $x^2-4x+1$.

I can solve it by writing $$a(x^{11}-x^4+1)^{15}-x^2+bx-c=(x^3+x)q(x)+x^2-4x+1$$

And solve the system of linear equations by looking at $x\in\{0,i,-i\}$ since those are the roots of $x^3+x$ making the term $(x^3+x)q(x)=0$ in the above expression. This leaves us with three simple linear equations where I got $(a,b,c)=(2,-6,1)$.

But what if the roots of the denominator weren't so easily obtainable because of a high degree, or the denominator had less unique roots than the unknown parameters? Then this method wouldn't work.

Are there other ways a problem like this could be solved relatively easily, if this was the case?

Or would that make things more complicated; what would be best way to solve something like that then?

Answer 1

Not sure if its easier, but here is another way:

Note we have $x^{11}-x^4+1 \equiv 1 \pmod x$ and $x^{11}-x^4+1 \equiv -x \pmod {x^2+1} $
So we have $(x^{11}-x^4+1)^{15} \equiv 1 \pmod x $ and $(x^{11}-x^4+1)^{15} \equiv x \pmod {x^2+1} $

Also as $(x^2+1)^{-1} \equiv 1 \pmod x$ and $x^{-1} \equiv -x \pmod {x^2+1}$, we apply CRT:

Thus, $\pmod {x^3+x}$ we must have $(x^{11}-x^4+1)^{15} \equiv (1)(x^2+1)(1) + (x)(x)(-x) = x^2+1-x^3 \equiv x^2+x+1$

$$\implies a(x^2+x+1) -x^2+bx-c \equiv x^2-4x+1$$ $$\iff (a-1)x^2+ (a+b)x+(a-c) \equiv x^2-4x+1$$ $\therefore a = 2, b = -6, c = 1$ is a solution.

Answer 2

If in $Q(x,a,b,c,\cdots)=d(x)q(x)+r(x)$ you express $q(x)$ by a polynomial with undetermined coefficients, then you get a $(n+1) \times (n+1)$ linear system, $n$ being the degree of $Q$, in the unknown $a,b,c, \cdots$ plus the coefficients of $q(x)$ which is of degree $n-$ degree of the denominator (call it $m$).

Then you can conclude easily by yourself.