Question

Find all the zeroes of the polynomial $p\left(x\right)=x^3+6-7x$ if one of its zeroes is $–3.$

Answer 1

Step 1: Obtain the factor of the cubic polynomial $p\left(x\right)=x^3+6-7x$:

Here,$–3$ is a zeroes of the polynomial $p\left(x\right)=x^3-7x+6.$

So, $x+3$is a factor of $p\left(x\right)$.

Step 2: Apply the long division method:

Now, we divide $p\left(x\right)$ by $\left(x+3\right),$we get

$\begin{array}{ccc}x& +& 3\end{array}\begin{array}{ccccc}{x}^2& -& 3x& +& 2\end{array}\begin{array}{cccccc}{x}^3& +& 0x^2& -& 7x& +\begin{array}{c}6\end{array}\\ x^3& +& 3x^2& & & \end{array}\begin{array}{cccccc}-& & -& & & \end{array}\begin{array}{cccccc}-& 3x^2& -& 7x& & \\ -& 3x^2& -& 9x& & \end{array}\begin{array}{cccccc}+& & +& & & \end{array}\begin{array}{cccccc}& & & 2x& +& 6\\ & & & 2x& +& 6\end{array}\begin{array}{cccccc}& & & -& & -\end{array}0$

Step 3: Apply the division algorithm to the polynomial $p\left(x\right)=x^3+6-7x$ and $x+3$:

As we know that

$\text{Dividend=Divisior ×Quotient+Remainder}$

$\begin{array}{rcl}& \Rightarrow & \mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}+3\right)({\mathrm{x}}^2-3\mathrm{x}+2)\\ & =& \left(\mathrm{x}+3\right)\left({\mathrm{x}}^2-2\mathrm{x}-\mathrm{x}+2\right)\\ & =& \left(\mathrm{x}+3\right)\left[\mathrm{x}\left(\mathrm{x}-2\right)-1\left(\mathrm{x}-2\right)\right]\\ & =& \left(\mathrm{x}+3\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-1\right)\end{array}$

As $p\left(x\right)=0$ when $x=-3,2,1$

Therefore, the zeroes of cubic polynomial are $2,1$ and $-3$.