# What are all the possible rational zeros for f(x)=4x^5+4x^2-9 and how do you find all zeros?

Dec 8, 2016

This quintic has no rational zeros. It has one irrational Real zero and $4$ non-Real complex ones.

#### Explanation:

$f \left(x\right) = 4 {x}^{5} + 4 {x}^{2} - 9$

By the rational root 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 $- 9$ and $q$ a divisor of the coefficient $4$ of the leading therm.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm \frac{9}{4} , \pm 3 , \pm \frac{9}{2} , \pm 9$

We find:

$f \left(1\right) = 4 + 4 - 9 = - 1 < 0$

$f \left(\frac{3}{2}\right) = 4 \left(\frac{243}{32}\right) + 4 \left(\frac{9}{4}\right) - 9 = \frac{243}{8} > 0$

So $f \left(x\right)$ has an irrational zero in $\left(1 , \frac{3}{2}\right)$

Looking at the derivative of $f \left(x\right)$ we find:

$f ' \left(x\right) = 20 {x}^{4} + 8 x = 4 x \left(5 {x}^{3} + 2\right)$

So $f \left(x\right)$ has turning points at $x = 0$ and $x = - \sqrt{\frac{2}{5}}$

$f \left(0\right) = - 9 < 0$

$f \left(- \sqrt{\frac{2}{5}}\right) = 4 {\left(\sqrt{\frac{2}{5}}\right)}^{2} \left(\frac{2}{5} + 1\right) - 9 = \frac{28}{5} {\left(\sqrt{\frac{2}{5}}\right)}^{2} - 9 < 6 - 9 < 0$

So the only Real zero of $f \left(x\right)$ is the irrational one in $\left(1 , \frac{3}{2}\right)$

Typically for a quintic, it is not possible to express the zeros in terms of radicals or elementary functions (including trigonometric, logarithmic or exponential functions).

The other $4$ Complex zeros will occur as two pairs of complex conjugates.

It is possible to find numerical approximations using algorithms such as Durand-Kerner.