Question

Let $f\left(x\right)=x^{135}+x^{125}-x^{115}+x^5+1$. If $f\left(x\right)$ is divided by $x^3-x$ then the remainder is some function of $x$ say $g\left(x\right)$. Find the value of $g\left(10\right)$.

Answer 1

$f\left(x\right)=x^{135}+x^{125}-x^{115}+x^5+1$

$f\left(x\right)=x(x-1)(x+1)q\left(x\right)+a{x}^2+ba+c$

$f\left(0\right)=C\Rightarrow C=1$ from i

$f\left(1\right)=3\Rightarrow a+b+c=3\Rightarrow a+b+1=3$

$a+b=2$----ii

$f(-1)=-1\Rightarrow a-b+c=-1$

$\Rightarrow a-b+1=-1$

$a-b=-2$

$ii-iii\Rightarrow 2a=0\Rightarrow a=0$ and $b=2$

$\therefore g\left(x\right)=2x+1\to g\left(10\right)=2\left(10\right)+1=21$