Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$

Question

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

Answer 1

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 $x=-1$, $$12=0+k(6)\implies k=2.$$

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

Answer 2

Proceeding from where I left off

$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$$$=\frac{x^{12}-1+2}{x+1}=\frac{x^{12}-1}{x+1}+\frac{2}{x+1}=P(x)+\frac{2}{x+1}$$where $P(x)$ is a polynomial since $x^{12}-1$ is divisible by $x+1$.

So $$1+x^2+x^4+x^6+....+x^{22}=P(x)\left(1+x+x^2+x^3+...+x^{11}\right)+\frac{2\left(1+x+x^2+x^3+...+x^{11}\right)}{x+1}$$$$=P(x)\left(1+x+x^2+x^3+...+x^{11}\right)+2\frac{x^{12}-1}{x^2-1}=P(x)\left(1+x+x^2+x^3+...+x^{11}\right)+2Q(x)$$where $Q(x)=1+x^2+x^4+x^6+x^8+x^{10}$

So answer should be $$2\left(1+x^2+x^4+x^6+x^8+x^{10}\right)$$

Answer 3

Upon applying: $\ fg\bmod fh\, =\, f(g\bmod h) =\, $ mod Dstributive Law with $\,z = x^2$

$\begin{align} &\ \,1+\cdots+z^{11}\,\bmod\, (1+\cdots+z^{5})(x+1)\\[.3em] =\ &(1+\cdots+z^{5})\big(1+\color{#c00}{z^6}\,\bmod x+1 \big)\\[.3em] =\ &(1+\cdots+z^{5})\big( 1 + \color{#c00}1\big)\, \ {\rm by}\ \color{#c00}{z^6}\equiv (x^2)^6\equiv \color{#c00}{1},\, \ {\rm by}\ \ x\equiv -1\!\!\!\pmod{\!x+1} \end{align}$