Question

Divide $(x^3-3x^2-x+3)$ by $(x+1)$and verify the division algorithm.

Answer 1

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

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

$g\left(x\right)=x+1$

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

$\begin{array}{ccc}x& +& 1\end{array}\begin{array}{ccccc}{x}^2& -& 4x& +& 3\end{array}\begin{array}{ccccccc}{x}^3& -& 3x^2& -& x& +& 3\\ x^3& +& x^2& & & & \\ -& -& & & & & \end{array}\begin{array}{}\end{array}\begin{array}{cccccc}-& 4x^2& -& x& +& 3\\ -& 4x^2& -& 4x& & \\ +& & +& & & \end{array}\begin{array}{ccc}3x& +& 3\\ 3x& +& 3\\ -& -& \end{array}\begin{array}{c}0\end{array}$

Therefore,

Here, Quotient = $\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^2-4\mathrm{x}+3$
And remainder = $\mathrm{r}\left(\mathrm{x}\right)=0$

Step 2: Verify the division algorithm:

Division Algorithm for polynomials states that:

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

$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\bullet \mathrm{q}\left(\mathrm{x}\right)+\mathrm{r}\left(\mathrm{x}\right),$

Where $\mathrm{r}\left(\mathrm{x}\right)=0$ or $degr\left(x\right)<degg\left(x\right).$
Then using division algorithm for polynomials we get;

$$\begin{gathered} q\left(x\right)\times g\left(x\right)+r\left(x\right)=(\mathrm{x}+1)({\mathrm{x}}^2-4\mathrm{x}+3)+0 \\ =\mathrm{x}({\mathrm{x}}^2-4\mathrm{x}+3)+1({\mathrm{x}}^2-4\mathrm{x}+3) \\ ={\mathrm{x}}^3-4{\mathrm{x}}^2+3\mathrm{x}+{\mathrm{x}}^2-4\mathrm{x}+3 \\ ={\mathrm{x}}^3-3{\mathrm{x}}^2-\mathrm{x}+3 \\ =p\left(x\right) \end{gathered}$$

Hence, the division algorithm is verified.