Question

On dividing $x^5-4x^3+x^2+3x+1$ by polynomial $g\left(x\right)$, the quotient and remainder are $(x^2-1)$ and $2$ respectively. Find $g\left(x\right)$.

• A. $g\left(x\right)=7x^2-x+9$
• B. $g\left(x\right)=x^3-3x+1$
• C. $g\left(x\right)=7x^3-3$
• D. $g\left(x\right)=7x^2-3x$

Answer 1

The correct option is B $g\left(x\right)=x^3-3x+1$

Let $q\left(x\right),g\left(x\right)$ and $r\left(x\right)$ be quotient, divisor and remainder respectively.

By remainder theorem,
$f\left(x\right)=q\left(x\right)g\left(x\right)+r\left(x\right)$
$\therefore x^5-4x^3+x^2+3x+1=(x^2-1)g\left(x\right)+2$
$\Rightarrow (x^2-1)g\left(x\right)=x^5-4x^3+x^2+3x-1$
Now,
$x^2-1)\overline{x^5-4x^3+x^2+3x-1}$ ( $x^3-3x+1=g\left(x\right)$
$\underset{–}{{\underset{-}{x}}^5\underset{+}{-}{x}^3}$
$-3x^3+x^2+3x-1$
$\underset{–}{\underset{+}{-}3x^3\underset{-}{+}3x}$
$x^2-1$
$\underset{–}{{\underset{-}{x}}^2\underset{+}{-}1}$
$0$
Hence, $q\left(x\right)=x^3-3x+1$