Question

Find the remainder when $x^{100}$ is divided by $x^2-3x+2$.

Answer 1

As per the reminder theory for any

polynomial equation

$p\left(x\right)=g\left(x\right).q\left(x\right)+r\left(x\right)\to \left(i\right)$

where $p\left(x\right),g\left(x\right)=$ polynomials

$q\left(x\right)=$ quotient

$r\left(x\right)=$ reminder

Assuming $r\left(x\right)=Ax+B$

here $p\left(x\right)=x^{100},g\left(x\right)=x^2-3x+2$

Simplifying $g\left(x\right)$

$g\left(x\right)=x^2-3x+2$

$=x^2-2x-x+2$

$=(x-2)(x-1)$

Substituting in eq${}^n$ (i)

$p\left(x\right)=(x-2)(x-1).q\left(x\right)+Ax+B$

taking $\left(x\right)=1$

$p\left(1\right)=(1-2)(1-1).q\left(1\right)+A\left(1\right)+B$

$1^{100}=0+A+B$

$A+B=1\to \left(i\right)$

taking $x=2$

$p\left(2\right)=(2-2)(2-1).q\left(x\right)+A\left(2\right)+B$

$2^{100}=0+2A+B$

$2A+B=2^{100}\to \left(ii\right)$

from eq${}^n$ (i) $B=1-A$ putting it in eq${}^n$ (i)

$2A+1-A=2^{100}$

$A=2^{100}-1$

now $B=1-2^{100}+1=2-2^{100}$

$\therefore B(2-2^{100})$

$\therefore$ Reminder $=Ax+B$

$=(2^{100}-1)-x+(2-2^{100})$