Question

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

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

Answer 1

The correct option is C $1+x^2+x^4+\dots .+x^{10}$

$\text{Let}P\left(x\right)=1+x^2+x^4+x^6+\dots .+x^{22}\text{and}Q\left(x\right)=1+x+x^2+x^3+\dots .+x^{11}P\left(x\right)=(1+x^2)(1+x^4)(1+x^4+x^8)(1-x^4+x^8)Q\left(x\right)=(1+x)(1+x^2)(1+x^4+x^8)\therefore \frac{P\left(x\right)}{Q\left(x\right)}=\frac{(1+x^4)(1-x^4+x^8)}{1+x}=\frac{1-x^4+x^8+x^4-x^8+x^{12}}{1+x}=\frac{1+x^{12}}{1+x}$
Remainder when $1+x^{12}$ is divided by $(1+x)$ is 2.
Therefore, the remainder of $P\left(x\right)$ divided by $Q\left(x\right)$ is
$=2(1+x^2)(1+x^4+x^8)$
$=2(1+x^2+x^4.....+x^{10})$