Question

Divide the polynomial $6x^3+11x^2-39x-65$ by the polynomial $g\left(x\right)=x^2-1+x$ Also find the quotient and remainder.

Answer 1

Step 1: Find the quotient and remainder by long division method:

As, $p\left(x\right)=6x^3+11x^2-39x-65$

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

Let the quotient and remainder be $\mathrm{q}\left(\mathrm{x}\right)$ and $\mathrm{r}\left(\mathrm{x}\right)$

$\begin{array}{ccccc}{x}^2& +& x& -& 1\end{array}\begin{array}{ccc}6x& +& 5\end{array}\begin{array}{cccccc}6x^3& +11x^2& -& 39x& -& 65\end{array}\begin{array}{ccccc}6x^3& +& 6x^2& -& 6x\\ -& -& & +& \end{array}\begin{array}{ccccc}5x^2& -& 33x& -& 65\\ 5x^2& +& 5x& -& 5\\ -& -& & +& \end{array}\begin{array}{cccc}-& 38x& -& 60\end{array}$

Clearly, the quotient = $q\left(x\right)=$ $(6x+5)$

and the remainder = $r\left(x\right)=$ $(-38x-60).$

Step 2: Verify the division algorithm:

If $\mathrm{f}\left(\mathrm{x}\right)$ and $\mathrm{g}\left(\mathrm{x}\right)$ are any two polynomials with $\mathrm{g}\left(\mathrm{x}\right)\ne 0,$

Then $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\times \mathrm{q}\left(\mathrm{x}\right)+\mathrm{r}\left(\mathrm{x}\right),$ where $\mathrm{r}\left(\mathrm{x}\right)=0$ or $\mathrm{deg}\mathrm{r}\left(\mathrm{x}\right)<\mathrm{deg}\mathrm{g}\left(\mathrm{x}\right).$

$$\begin{gathered} q\left(x\right)\times g\left(x\right)+r\left(x\right)=(6x+5)(x^2+x-1)+(-38x-60) \\ =6x^3+11x^2-x-5-38x-60 \\ =6x^3+11x^2-39x-65 \\ =p\left(x\right) \end{gathered}$$

Hence, the division algorithm is verified.