Question

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

• A. $0$
• B. $2$
• C. $1+x^2+x^4+...+x^{10}$
• D. $2(1+x^2+x^4+....+x^{10})$

Answer 1

The correct option is D $2(1+x^2+x^4+....+x^{10})$

$\left({\sum }_{n=0}^N{x}^n\right)=\left(\frac{x^{N+1}-1}{x-1}\right)$

$\Rightarrow Dividend=\left(\frac{x^{24}-1}{x^2-1}\right)$

$Divisor=\left(\frac{x^{12}-1}{x-1}\right)$

Now,

$\left(\frac{x^{24}-1}{x^2-1}\right)=\left(\frac{x^{12}-1}{x-1}\right)\left(\frac{x^{12}-1+2}{x+1}\right)=\left(\frac{{(x^{12}-1)}^2}{x^2-1}\right)+2\left(\frac{x^{12}-1}{x^2-1}\right)$

$\Rightarrow Remainder=2(1+x^2+x^4...+x^{10})$