<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] Upon applying: $\ fg\bmod fh\, =\, f(g\bmod h) =\, $ mod Dstributive Law with $\,z = x^2$
$\begin{align}
&\ \,1+\cdots+z^{11}\,\bmod\, (1+\cdots+z^{5})(x+1)\\[.3em]
=\ &(1+\cdots+z^{5})\big(1+\color{#c00}{z^6}\,\bmod x+1 \big)\\[.3em]
=\ &(1+\cdots+z^{5})\big( 1 + \color{#c00}1\big)\, \ {\rm by}\   \color{#c00}{z^6}\equiv (x^2)^6\equiv \color{#c00}{1},\, \ {\rm by}\ \ x\equiv -1\!\!\!\pmod{\!x+1}
\end{align}$
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