Question

Find the quotient $q\left(x\right)$ and remainder $r\left(x\right)$ of the following when $f\left(x\right)$ is divided by $g\left(x\right)$.
$f\left(x\right)=x^3-3x^2-x+3$;
$g\left(x\right)=x^2-4x+3$

• A. $q\left(x\right)=x-1$ and $r\left(x\right)=0$
• B. $q\left(x\right)=x+1$ and $r\left(x\right)=0$
• C. $q\left(x\right)=2x+1$ and $r\left(x\right)=0$
• D. $q\left(x\right)=2x-1$ and $r\left(x\right)=0$

Answer 1

The correct option is B $q\left(x\right)=x+1$ and $r\left(x\right)=0$

Consider the polynomial $f\left(x\right)=x^3-3x^2-x+3$ and factorise it as follows:

$f\left(x\right)=x^3-3x^2-x+3=x^2(x-3)-1(x-3)=(x^2-1)(x+3)=(x+1)(x-1)(x-3)$

Therefore, $f\left(x\right)=(x+1)(x-1)(x-3)$

Now consider the polynomial $g\left(x\right)=x^2-4x+3$ and factorise it as follows:

$g\left(x\right)=x^2-4x+3=x^2-3x-x+3=x(x-3)-1(x-3)=(x-1)(x-3)$

Therefore, $g\left(x\right)=(x-1)(x-3)$

We know that, $f\left(x\right)$ = $q\left(x\right)g\left(x\right)+r\left(x\right)$.

Now divide $f\left(x\right)$ by $g\left(x\right)$ to get $q\left(x\right)$:

$q\left(x\right)=\frac{(x-1)(x+1)(x-3)}{(x-1)(x-3)}=x+1$

Since $f\left(x\right)$ is divisible by $g\left(x\right)$, therefore, the remainder $r\left(x\right)=0$.

Hence, the quotient $q\left(x\right)=x+1$ and remainder $r\left(x\right)=0$.