Answer
The zeros of the given polynomial are $3,-2$ and $1.$
$f\left( x \right)={{x}^{3}}-2{{x}^{2}}-5x+6$
Work Step by Step
First find the possible rational zeros of the provided function as given below:
$\begin{align}
& \text{Possible rational zeros }=\frac{\text{Factors of the constant term, 6}}{\text{Factors of the leading coefficient, 1}} \\
& =\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1} \\
& =\pm 1,\pm 2,\pm 3,\pm 6
\end{align}$
Then, find the possible rational zeros of the given function. Divide the factors of the constant term 6 of $f\left( x \right)$ by the factors of the leading coefficient 1 of $f\left( x \right)$. Factors of the constant term 6 are $\pm 1,\pm 2,\pm 3,\pm 6$. Factors of the leading coefficient 1 is $\pm 1$. After that, divide each factor of 6 by the factor of 1 to get the possible rational zeros, that is $\pm 1,\pm 2,\pm 3,\pm 6$.
Now, use the first considered possible rational zero 1 as the divisor for synthetic division. And write the coefficient of $f\left( x \right)$ in the first row in descending order of the power of $x.$ Perform synthetic division as follows:
$\begin{align}
& \underline{\begin{align}
& \text{coefficient of } \\
& f\left( x \right)={{x}^{3}}-2{{x}^{2}}-5x+6 \\
& \,\Downarrow \\
& \left. {\underline {\,
1 \,}}\! \right| 1-2-5\,\,\,\,\,\,\,\,\,\,6 \\
& 1-1\,-6 \\
\end{align}} \\
& \,1-1-6\,\,\,\,\,\,\,\,\,\,0\,\leftarrow \,\text{reminder} \\
\end{align}$
Since the reminder is zero, 1 is a zero of $f\left( x \right)={{x}^{3}}-2{{x}^{2}}-5x+6.$ Therefore, $\left( x-1 \right)$ is a factor of the given $f\left( x \right)$. To find the other factor, divide $f\left( x \right)$ by $\left( x-1 \right)$ using normal division until the reminder becomes zero as follows:
$\begin{align}
& {{x}^{2}}-x-6 \\
& x-1\left| \!{\overline {\,
{{x}^{3}}-2{{x}^{2}}-5x+6 \,}} \right. \\
& \underline{{{x}^{3}}-x{{}^{2}}\,\,} \\
& \underline{\begin{align}
& -{{x}^{2}}-5x\, \\
& -{{x}^{2}}+x\, \\
\end{align}} \\
& -6x+6 \\
& \,\underline{-6x+6} \\
& 0 \\
& \\
\end{align}$
So, the quotient $\left( {{x}^{2}}-x-6 \right)$ is also a factor of the given $f\left( x \right)$. $\left( {{x}^{2}}-x-6 \right)$ We can factor as follows:
$\left( {{x}^{2}}-x-6 \right)=\left( x-3 \right)\left( x+2 \right)$
Thus,
$\begin{align}
& f\left( x \right)={{x}^{3}}-2{{x}^{2}}-5x+6 \\
& =\left( x-1 \right)\left( x-3 \right)\left( x+2 \right)
\end{align}$
To find zeros of $f\left( x \right)$, equate the function to zero. That means:
$\begin{align}
& f\left( x \right)=0 \\
& {{x}^{3}}-2{{x}^{2}}-5x+6=0 \\
& \left( x-1 \right)\left( x-3 \right)\left( x+2 \right)=0
\end{align}$
This means $x=1\text{ or 3 or}-2.$
