Question

Find all the zeroes of $x^3-7x+6,$if one of its zero is $-3.$

Answer 1

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

Here the zero of the polynomial is $-3$.

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

Step 2: Find the quotient using long division method:

Let $g\left(x\right)=x+3$.

Divide the polynomial $p\left(x\right)$ by $g\left(x\right)$.

$\begin{array}{ccc}x& +& 3\end{array}\begin{array}{cccccc}{x}^2& -& 3x& +& 2& \end{array}\begin{array}{cccccc}{x}^3& +& 0x^2& -& 7x& +\begin{array}{c}6\end{array}\\ x^3& +& 3x^2& & & \\ \left(-\right)& \left(-\right)& & & & \end{array}\begin{array}{cccccc}-& 3x^2& -& 7x& +& 6\\ -& 3x^2& -& 9x& & \\ \left(+\right)& & \left(+\right)& & & \end{array}\begin{array}{}\end{array}\begin{array}{cccc}& 2x& +& 6\\ & 2x& +& 6\\ & \left(-\right)& \left(-\right)& \end{array}0$

Since the remainder is $0$, $g\left(x\right)=x+3$ is a factor of $p\left(x\right)=x^3-7x+6$.

Step 3: Use division algorithm to find the other zeroes:

Formula:

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

$\Rightarrow x^3-7x+6=\left(x+3\right)\left(x^2-3x+2\right)$.

Consider,

$$\begin{gathered} \Rightarrow x^2-3x+2=0 \\ \Rightarrow x^2-x-2x+2=0 \\ \Rightarrow x\left(x-1\right)-2\left(x-1\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)=0 \end{gathered}$$

Either $x-1=0$ or $x-2=0$, that is, $x=1$ or $x=2$.

Therefore, the other zeroes are $1$ and $2$.