Question

The remainder when $x^{100}$ is divided by $x^2-3x+2$ is :

• A. $(2^{100}-1)x+(-2^{100}+2)$
• B. $(2^{100}+1)x+(-2^{100}-2)$
• C. $(2^{100}-1)x+(-2^{100}-2)$
• D. None

Answer 1

The correct option is A $(2^{100}-1)x+(-2^{100}+2)$

$Here$ $the$ $remainder$ $is$ $assumed$ $to$ $be$ $a$ $linear$ $since$ $the$ $divisor$ $is$ $a$ $quadratic$ $one.$
$Thus$ $remainder$ $=ax+b$
$So,$ $x^{100}$ $=Quotient\times (x^2-3x+2)$ $+$ $(ax+b)$
$Putting$ $x=1,$ $we$ $have$ $1=0+(a+b)$ $and$
$putting$ $x=2,$ $we$ $have$ $2^{100}=0+(2a+b)$
$Thus$ $solving$ $simultaneously,$ $a=2^{100}-1$ $and$ $b=2-2^{100}$
$\therefore$ $remainder$ $=(2^{100}-1)x+(2-2^{100})$