# What are all the possible rational zeros for f(x)=2x^3+x^2+8x+4 and how do you find all zeros?

Jun 23, 2017

One rational zero at $x = - \frac{1}{2}$; two complex roots at $x = \pm \sqrt{2} i$

#### Explanation:

The Rational Root Theorem tells us that given a function of the form:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{x - 1} + \ldots + {a}_{1} x + {a}_{0}$, with ${a}_{n \ldots 0} \in \mathbb{Z}$
all rational zeroes will be of the form:
$\textcolor{w h i t e}{\text{XXX}} \frac{q}{p}$ where
color(white)("XXXXXX") p " is a positive integer factor of "a_n" and
color(white)("XXXXXX") q" is an integer factor of "a_0

We can test all such possibilities (I would suggest using a spread sheet as below) for $2 {x}^{3} + {x}^{2} + 8 x + 4$: This gives us that there is only one rational zero at
$\textcolor{w h i t e}{\text{XXX}} x = - \frac{1}{2}$
which implies
$\textcolor{w h i t e}{\text{XXX}} \left(x + \frac{1}{2}\right)$ is a factor of $2 {x}^{3} + {x}^{2} + 8 x + 4$

Using long division or synthetic division (with $\left(x + \frac{1}{2}\right)$ as the divisor) we find:
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} + {x}^{2} + 8 x + 4 = \left(x + \frac{1}{2}\right) \left(2 {x}^{2} + 4\right)$

For zeroes of the given function we have:
{: ("either ",(x+1/2)=0," or ",(2x^2+4)=0), (,rarr x=-1/2,,rarr x^2+2=0), (,"which we already knew",,rarr x^2=-2), (,,,rarr x=sqrt(2)i) :}