# What are all the possible rational zeros for f(x)=5x^4+32x^2-21?

Jan 8, 2017

The "possible" rational zeros are:

$\pm \frac{1}{5} , \pm \frac{3}{5} , \pm 1 , \pm \frac{7}{5} , \pm 3 , \pm \frac{21}{5} , \pm 7 , \pm 21$

The actual zeros are:

$\pm \frac{\sqrt{15}}{5} \text{ }$ and $\text{ } \pm \sqrt{7} i$

#### Explanation:

Given:

$f \left(x\right) = 5 {x}^{4} + 32 {x}^{2} - 21$

If we apply the rational root theorem directly, 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 $- 21$ and $q$ a divisor of the coefficient $5$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{5} , \pm \frac{3}{5} , \pm 1 , \pm \frac{7}{5} , \pm 3 , \pm \frac{21}{5} , \pm 7 , \pm 21$

Note also that when $\left\mid x \right\mid \ge 1$, then:

$f \left(x\right) \ge 5 + 32 - 21 = 16$

So the only possible rational zeros are:

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

and since all the terms in $x$ are of even degree, if ${x}_{1}$ is a zero then so is $- {x}_{1}$.

So we only need to check $x = \frac{1}{5}$ and $x = \frac{3}{5}$...

$f \left(\frac{1}{5}\right) = 5 \left(\frac{1}{625}\right) + 32 \left(\frac{1}{25}\right) - 21 = \frac{1 + 160 - 2625}{125} = - \frac{2464}{125}$

$f \left(\frac{3}{5}\right) = 5 \left(\frac{81}{625}\right) + 32 \left(\frac{9}{25}\right) - 21 = \frac{81 + 1440 - 2625}{125} = - \frac{1104}{125}$

So $f \left(x\right)$ has no rational zeros.

$\textcolor{w h i t e}{}$
We can factor $f \left(x\right)$ as a quadratic in ${x}^{2}$ using an AC method:

Look for a pair of factors of $A C = 5 \cdot 21 = 105$ with difference $B = 32$

The pair $35 , 3$ works.

Use this pair to split the middle term and factor by grouping:

$5 {x}^{4} + 32 {x}^{2} - 21 = \left(5 {x}^{4} + 35 {x}^{2}\right) - \left(3 {x}^{2} + 21\right)$

$\textcolor{w h i t e}{5 {x}^{4} + 32 {x}^{2} - 21} = 5 {x}^{2} \left({x}^{2} + 7\right) - 3 \left({x}^{2} + 7\right)$

$\textcolor{w h i t e}{5 {x}^{4} + 32 {x}^{2} - 21} = \left(5 {x}^{2} - 3\right) \left({x}^{2} + 7\right)$

Hence we can see that the zeros of $f \left(x\right)$ are actually:

$x = \pm \sqrt{\frac{3}{5}} = \pm \frac{\sqrt{15}}{5} \text{ }$ and $\text{ } x = \pm \sqrt{7} i$