<s>[INST] $x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:

If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$

In these types of questions generally I follow the following approach:

Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$

But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:

 Option (B)

 [/INST] You want to divide by $x^3+x = x(x^2+1)$, hence by the Chinese remainder theorem it is enough to check the remainders $\pmod{x}$ (which is obviously zero) and $\pmod{x^2+1}$. This remainder is simply given by setting $x^2\equiv -1$, such that
$$ x^{10}+x^{11}+\ldots+x^{19}+x^{20} \equiv (-1-x+1+x)+(-1-x+1+x)+(-1-x+1+x)\color{red}{-x} $$
and option $(B)$ is apparent now.
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