# What are all the possible rational zeros for f(x)=4x^5-2x^4+30x^3-15x^2+50x-25?

Sep 12, 2016

See explanation...

#### Explanation:

$f \left(x\right) = 4 {x}^{5} - 2 {x}^{4} + 30 {x}^{3} - 15 {x}^{2} + 50 x - 25$

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 $25$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1 , \pm \frac{5}{4} , \pm \frac{5}{2} , \pm 5 , \pm \frac{25}{4} , \pm \frac{25}{2} , \pm 25$

In fact, note that the signs of the coefficients alternate, with no powers of $x$ missing. Hence (by Descartes rule of signs) there are $1$, $3$ or $5$ positive Real zeros and no negative Real zeros.

So the only possible rational zeros are:

$\frac{1}{4} , \frac{1}{2} , 1 , \frac{5}{4} , \frac{5}{2} , 5 , \frac{25}{4} , \frac{25}{2} , 25$

Trying each in turn, we find:

$f \left(\frac{1}{2}\right) = 4 \left(\frac{1}{32}\right) - 2 \left(\frac{1}{16}\right) + 30 \left(\frac{1}{8}\right) - 15 \left(\frac{1}{4}\right) + 50 \left(\frac{1}{2}\right) - 25$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = \frac{1 - 1 + 30 - 30 + 200 - 200}{8} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$4 {x}^{5} - 2 {x}^{4} + 30 {x}^{3} - 15 {x}^{2} + 50 x - 25 = \left(2 x - 1\right) \left(2 {x}^{4} + 15 {x}^{2} + 25\right)$

We can factor the remaining quartic as a quadratic in ${x}^{2}$

$2 {x}^{4} + 15 {x}^{2} + 25 = 2 {x}^{4} + 10 {x}^{2} + 5 {x}^{2} + 25$

$\textcolor{w h i t e}{2 {x}^{4} + 15 {x}^{2} + 25} = 2 {x}^{2} \left({x}^{2} + 5\right) + 5 \left({x}^{2} + 5\right)$

$\textcolor{w h i t e}{2 {x}^{4} + 15 {x}^{2} + 25} = \left(2 {x}^{2} + 5\right) \left({x}^{2} + 5\right)$

Hence zeros:

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