# 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

• you are off to a good start. Now just use the standard polynomial division. – dezdichado Apr 2 at 4:23
• As @dezdichando said you can use the standard polynomial division, but there is a much better way. Using the Polynomial remainder theorem, you can find an example here or [wikipedia][2] [2]: en.wikipedia.org/wiki/Polynomial_remainder_theorem – Baby desta Apr 2 at 4:37

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

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

• We can avoid solving equations by using distributivity - see my answer. – Bill Dubuque Apr 2 at 22:50

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

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}