$x^{10}+x^{11}+\dots+x^{20}$ 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)


We want to find the remainder from the division:$$\frac{x^{20} + x^{19} + \ldots + x^{11} + x^{10}}{x^3 + x}$$
Now note that $$ (x^3 + x)(x^{n+1} + x^n ) = x^{n+4} + x^{n+3} + x^{n+2} + x^{n+1}, $$
so that
$$ \frac{x^{20} + x^{19} + \ldots + x^{11} + x^{10}}{x^3 + x} = x^{17} + x^{16} + x^{13} + x^{12} + \frac{x^{12} + x^{11} + x^{10}}{x^3 + x} $$
and using long division you can show that the remainder from the final term is  $-x.$
Peter's method is more efficient though.
