Question

Divide the polynomial $\mathrm{p}\left(\mathrm{x}\right)$ by the polynomial $\mathrm{g}\left(\mathrm{x}\right)$ and find quotient and remainder for the following:
$\mathrm{p}\left(\mathrm{x}\right)=4{\mathrm{x}}^3+5{\mathrm{x}}^2+6;\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}-2$

Answer 1

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

As, $p\left(x\right)=4{\mathrm{x}}^3+5{\mathrm{x}}^2+6$

$g\left(x\right)=x-2$

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

$\begin{array}{ccccc}& & x& -& 2\end{array}\begin{array}{ccc}4x^2& +& 13x\end{array}\begin{array}{cccccc}4x^3& +5x^2& +& 0x& +& 6\end{array}\begin{array}{ccccc}4x^3& -& 8x^2& +& 0x\\ -& +& & -& \end{array}\begin{array}{ccccc}+13x^2& -& 0x& +& 6\\ +13x^2& -& 26x& +& 0\\ -& +& & +& \end{array}\begin{array}{cccc}+& 26x& +& 6\end{array}$

Clearly, the quotient $=q\left(x\right)=4x^2+13x$

and the remainder $=r\left(x\right)=26x+6$

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 $\mathrm{deg}\mathrm{r}\left(\mathrm{x}\right)<\mathrm{deg}\mathrm{g}\left(\mathrm{x}\right).$

$$\begin{gathered} q\left(x\right)\times g\left(x\right)+r\left(x\right)=(x-2)(4x^2+13x)+(26x+6) \\ =4x^3+5x^2-26x+26x+6 \\ =4x^3+5x^2+6 \\ =p\left(x\right) \end{gathered}$$

Hence, the division algorithm is verified.