Question

please anyone solve this​

Answer 1

Method

Topic- Polynomial division

If we divide a polynomial by another polynomial of degree n $(n\geq 1)$, the degree of the remainder is at most n-1. It is because the polynomial cannot divide anything below its degree. Here $n=2$, so the remainder will be a linear polynomial $ax+b$.

Solution

Let the remainder be $r(x)=ax+b$.

Since it is a division, $f(x)=p(x)q(x)+r(x)$

Where

• $f(x)$ dividend
• $p(x)$ divisor
• $q(x)$ quotient
• $r(x)$ remainder

Given the dividend and divisor,

$\implies x^{100}-2x+1=(x+1)(x-1)q(x)+ax+b$

Substitute $x=1$ or $x=-1$. The following equations are obtained.

$x=1 \implies 1-2+1=0q(1)+a+b$

$x=-1\implies 1+2+1=0q(-1)-a+b$

So,

$\begin{cases} & a+b=0 \\ & -a+b=4 \end{cases}\implies a=-2,b=2$

Therefore the remainder must be $-2x+2$.