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


Here $x$ is common, hence we can "cancel" it to get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19}$ & $x^2+1$
By your method, let this be $f(x) = (x^2+1)g(x)+ax+b$
At $x=i$, we get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19} = 0 + ai+b$
At $x=-i$, we get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19} = 0 - ai+b$
LHS in both can be calculated by Geometric Progression.
We will end up with $2$ Simultaneous Equations having complex co-efficients.
Solving that will give the required reminder.
Alternately, we know that $x^4 = 1$ when $x=i$ or $x=-i$.
With that, we can reduce the given Equations:
$x^1+x^{2}+x^{3}+x^{0}+ \cdots +x^{3} = 0 + ai+b$
$x^1+x^{2}+x^{3}+x^{0}+ \cdots +x^{3} = 0 - ai+b$
$(+i)+(-1)+(-i)+(+1)+ \cdots +(-i) = 0 + ai+b$
$(-i)+(-1)+(+i)+(+1)+ \cdots +(+i) = 0 - ai+b$
$(0i)+(-1) = 0 + ai+b$
$(0i)+(-1) = 0 - ai+b$
Both give $(a,b)=(0,-1)$
Plugging into the original, before we did the "cancellation", we get $0x^2-x = -x$
