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 $g\left(x\right)$.
$p\left(x\right)=x^6+x^4-x^2-1$;
$g\left(x\right)=x^3-x^2+x-1$

• A. $q\left(x\right)=2x^3+x^2+x-1$ and $r\left(x\right)=0$
• B. $q\left(x\right)=x^3+x^2+x-1$ and $r\left(x\right)=0$
• C. $q\left(x\right)=2x^3+x^3+x+1$ and $r\left(x\right)=0$
• D. $q\left(x\right)=x^3+x^2+x+1$ and $r\left(x\right)=0$

Answer 1

The correct option is D $q\left(x\right)=x^3+x^2+x+1$ and $r\left(x\right)=0$

Consider the polynomial $f\left(x\right)=x^6+x^4-x^2-1$ and factorise it as follows:

$f\left(x\right)=x^6+x^4-x^2-1=x^4(x^2+1)-1(x^2+1)=(x^4-1)(x^2+1)=(x^2-1)(x^2+1)(x^2+1)$

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

Therefore, $f\left(x\right)=(x-1)(x+1)(x^2+1{)}^2$

Now consider the polynomial $g\left(x\right)=x^3-x^2+x-1$ and factorise it as follows:

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

Therefore, $g\left(x\right)=(x-1)(x^2+1)$

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 $q\left(x\right)$:

$q\left(x\right)=\frac{(x-1)(x+1)(x^2+1{)}^2}{(x-1)(x^2+1)}=(x+1)(x^2+1)=x^3+x^2+x+1$

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

Hence, the quotient $q\left(x\right)=x^3+x^2+x+1$ and remainder $r\left(x\right)=0$.