Here is a list of all possible rational zeroes for the polynomial.

\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 1}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 1}} & = \pm 6 & \hspace{0.25in}\frac{{ \pm 9}}{{ \pm 1}} & = \pm 9 & \hspace{0.25in}\frac{{ \pm 18}}{{ \pm 1}} & = \pm 18\\ & \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 2}} & = \frac{{ \pm 3}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 2}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 9}}{{ \pm 2}} & = \frac{{ \pm 9}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 18}}{{ \pm 2}} & = \pm 9\end{align*}\]

So, we have a total of 18 possible zeroes for the polynomial.

Show Step 3

We now need to start the synthetic division work. We’ll start with the “small” integers first.

\[\,\,\begin{array}{r|rrrl} {} & 2 & -13 & 3 & 18 \\ \hline -1 & 2 & -15 & 18 & 0=f\left( -1 \right)=0!! \\ \end{array}\]

Okay we now know that \(x = - 1\) is a zero and we can write the polynomial as,

\[f\left( x \right) = 2{x^3} - 13{x^2} + 3x + 18 = \left( {x + 1} \right)\left( {2{x^2} - 15x + 18} \right)\] Show Step 4

We could continue with this process however, we have a quadratic for the second factor and we can just factor this so the fully factored form of the polynomial is,

\[f\left( x \right) = 2{x^3} - 13{x^2} + 3x + 18 = \left( {x + 1} \right)\left( {2x - 3} \right)\left( {x - 6} \right)\] Show Step 5

From the fully factored form we get the following set of zeroes for the original polynomial.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = \frac{3}{2}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = 6}}\]