$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ 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.
 A: 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.
A: $x^3+x=x(x+i)(x-i)$
Remainder of $P(x)/(x^3+x)$ is a (at most) second degree polynomial 
$R(x)=a x^2+bx+c$
We have $P(i)=-1+2i$ and $R(i)=-a + i b + c$ so $-a + i b + c=-1+2i$
then $P(-i)=-1-2i$ and $R(-i)=-a - i b + c$ so $-a - i b + c=-1-2i$
and $P(0)=1$ and $R(0)=c$
So $c=1$. Substitute in the first two equations
$-a+ib+1=-1+2i;\;-a-ib+1=-1-2i$
add the two equations $-2a+2=-2$ then $a=2$
substitute and get $b=2$
Therefore the remainder is $R(x)=2x^2+2x+1$
Hope this helps
A: Applying: $\  fg\bmod fh\  =\ f\:(\,g\bmod h)\, =\, $ mod Distributive Law makes it very easy
$\quad \begin{align}P\!-1\,\bmod\, {x^{\large 3}\!+x}\, &=\, x\left(\dfrac{P\!-\!1}{x}\bmod\, \color{#c00}{x^{\large 2}+\,1}\right)\\[.3em]
&=\,x\, (\,2x+2\,)\, \  {\rm by} \, \ \color{#c00}{x^{\large 2}\!\equiv -1}\,\Rightarrow\, x^{\large 49}\!\equiv x(x^{\large 4})^{\large 12}\!\equiv x\,1^{\large 12}\!\equiv x\ \ {\rm etc}\\[.3em]
\end{align}$
A: Reduce the degree of polynomial using $x^3=-x$ from $d=100$ to $d=2$.
A: $P(x)=x^{100}+x^{50}-2x^4-x^3+x+1$
Let's divide $x^{100}+x^{50}$ by $x^3+x$
-> $Q=x^{97},r=-x^{98}+x^{50}$
Divide $-x^{98}+x^{50}$ by $x^3+x$
-> $Q=-x^{95}, r=x^{96}+x^{50}$
Divide $x^{96}+x^{50}$ by $x^3+x$
-> $Q=x^{93}, r=-x^{94}+x^{50}$
...
Can you see a pattern?
...
The pattern ends here:
Divide $x^{52}+x^{50}$ by $x^3+x$
-> $Q=x^{49}, r=0$

Now return to the original polynomial (without the first two term):
$-2x^4-x^3+x+1$
And do polynomial division as you do normally!
A: 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?
