Question

Let $r\left(x\right)$ be the remainder when the polynomial $x^{135}+x^{125}–x^{115}+x^5+1$ is divided by $x^3–x$. Then

• A. $r\left(x\right)$ is the zero polynomial
• B. $r\left(x\right)$ is a nonzero constant
• C. degree of $r\left(x\right)$ is one
• D. degree of $r\left(x\right)$ is two

Answer 1

The correct option is C degree of $r\left(x\right)$ is one

We can write
$x^{135}+x^{125}–x^{115}+x^5+1=A(x^3-x)+B{x}^2+Cx+D$
Putting $x=0\Rightarrow D=1$
Putting $x=1\Rightarrow B+C=2$.........(i)
Putting $x=-1\Rightarrow B-C=-2$.........(ii)
Using equation (i) and (ii),
$B=0,C=2$
$\therefore x^{135}+x^{125}–x^{115}+x^5+1=A(x^3-x)+2x+1$
Hence $r\left(x\right)=2x+1$