# What are all the possible rational zeros for f(x)=2x^4-6x^2+5x-15 and how do you find all zeros?

Aug 11, 2017

"Possible" rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm \frac{15}{2} , \pm 15$

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{4} - 6 {x}^{2} + 5 x - 15$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 15$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm 3 , \pm 5 , \pm \frac{15}{2} , \pm 15$

With the function as specified, none of these is a zero.

If we graph $f \left(x\right)$ then it looks like this:

graph{2x^4-6x^2+5x-15 [-5, 5, -35, 25]}

There are two irrational zeros, one somewhere in the interval $\left(- \frac{5}{2} , - 2\right)$, the other in the interval $\left(\frac{3}{2} , 2\right)$.

Typically for a quartic, these are possible but somewhat messy to find algebraically. If you would really like to see a solution, please request in the comments.

I suspect a typo in the question.

Consider the function:

${f}_{1} \left(x\right) = 2 {x}^{3} - 6 {x}^{2} + 5 x - 15$

The rational roots theorem suggests the same possible rational roots, but this time the graph looks like this:

graph{2x^3-6x^2+5x-15 [-5, 5, -35, 25]}

There is one real rational root at $x = 3$:

${f}_{1} \left(3\right) = 2 {\left(\textcolor{b l u e}{3}\right)}^{3} - 6 {\left(\textcolor{b l u e}{3}\right)}^{2} + 5 \left(\textcolor{b l u e}{3}\right) - 15$

$\textcolor{w h i t e}{{f}_{1} \left(3\right)} = 54 - 54 + 15 - 15$

$\textcolor{w h i t e}{{f}_{1} \left(3\right)} = 0$

We can also factor ${f}_{1} \left(x\right)$ by grouping:

$2 {x}^{3} - 6 {x}^{2} + 5 x - 15 = \left(2 {x}^{3} - 6 {x}^{2}\right) + \left(5 x - 15\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 6 {x}^{2} + 5 x - 15} = 2 {x}^{2} \left(x - 3\right) + 5 \left(x - 3\right)$

$\textcolor{w h i t e}{2 {x}^{3} - 6 {x}^{2} + 5 x - 15} = \left(2 {x}^{2} + 5\right) \left(x - 3\right)$

From this we can see the real zero $x = 3$ and find the complex zeros:

$x = \pm \sqrt{\frac{5}{2}} i = \pm \sqrt{\frac{10}{4}} i = \pm \frac{\sqrt{10}}{2} i$