$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:

$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$

What is the remainder of $\frac{P(X)}{x^3+x}$?

I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^3+x$ is not linear. Could I have some hints on how to approach this? Thank you.
We have
$$
P(x) = Q(x)(x^3 + x) + r(x)
$$
where $Q(x)$ and $r(x)$ are polynomials, and $r(x)$ has degree at most $2$. Now, considering that $Q(x)(x^3 + x)$ has roots at $x = -i, 0$ and $i$, we can use this to get information about $r(x)$, namely that $r(\pm i) = P(\pm i)$ and $r(0) = P(0)$. These three points together with what we know about the degree of $r$ is enough to find $r$ exactly.
The remainder is
$$\frac{x+2}{x^2+1}\:+\:\frac 1x\qquad\text{or}\quad2x^2+2x+1\;\;\text{respectively}\\[3ex]$$
obtained by taking a lazy approach: Feeding the command

apart((x**100+x**50-2*x**4-x**3+x+1)/(x**3+x))


into http://live.sympy.org/ ,
thus asking Python to go through the long division in short time,
yields
$$\frac{P(x)}{x(x^2+1)}\;=\;\sum_{m=48}^{24}(-1)^mx^{2m+1}-2x-1+\frac 1x+\frac{x+2}{x^2+1}\\[3ex]$$

Note that the $\,\sum\,$ equals $(x^{100}+x^{50})/(x^3+x)=x^{49}(x^{50}+1)/(x^2+1)\,$, whence no contribution to the remainder.

... did I fail to address the issue in question?