Paul's Online Notes
Home / Algebra / Polynomial Functions / Finding Zeroes of Polynomials
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 5.4 : Finding Zeroes of Polynomials

1. Find all the zeroes of the following polynomial.

$f\left( x \right) = 2{x^3} - 13{x^2} + 3x + 18$

Show All Steps Hide All Steps

Start Solution

We’ll need all the factors of 18 and 2.

\begin{align*}18 & : \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\\ 2 & : \pm 1,\,\, \pm 2\end{align*} Show Step 2

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.

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