Question

Find the remainder obtained when $x^{2007}$ is divisible by $x^2-1$.

• A. $x^2$
• B. $x$
• C. $x+1$
• D. $-x$

Answer 1

The correct option is B $x$

Here the remainder is assumed to be a linear since the divisor is a quadratic one.

Thus remainder = $ax+b$

So, $x^{2007}$ = Quotient$\times (x^2-1)+(ax+b)$

Putting $x=1$, we have $1=0+(a+b)$ and

putting $x=-1$, we have $-1^{2007}=-1=0+(-a+b)$

Thus solving simultaneously, $b=0$ and $a=1$

$\therefore$ remainder = $x+0=x$