<s>[INST] Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$ 
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$
Dont know how to proceed from here
 [/INST] Assume $\deg Q(x)=11$ and $\deg R(x)\le10$, with
\begin{align*}
(1+x^2+\cdots+x^{22})=(1+x+\cdots+x^{11})Q(x)+R(x).\tag 1
\end{align*}
Multiply $(x^2-1)$ on both sides,
\begin{align*}
\color{blue}{(x^{12}-1)}(x^{12}+1)&=\color{blue}{(x^{12}-1)}(x+1)Q(x)+(x^2-1)R(x)
\end{align*}
implies $\color{blue}{(x^{12}-1)}\mid(x^2-1)R(x)$. Since $\deg (x^2-1)R(x)\le12$, we have
\begin{align*}
R(x)=k\cdot\frac{x^{12}-1}{x^2-1}=k(1+x^2+\cdots +x^{10}),\qquad k\in\Bbb R. \tag 2
\end{align*}
Put $(2)$ into $(1)$ and substitute $x=-1$,
$$12=0+k(6)\implies k=2.$$

$$\therefore R(x)=2(1+x^2+\cdots +x^{10}).$$

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