# What are all the possible rational zeros for f(x)=2x^4-13x^3-34x^2+65x+120 and how do you find all zeros?

Aug 3, 2017

"Possible" rational zeros:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm \frac{5}{2} , \pm 3 , \pm 4 , \pm 5 , \pm 6 , \pm \frac{15}{2} , \pm 8 , \pm 10 , \pm 12 , \pm 15 , \pm 20 , \pm 24 , \pm 30 , \pm 40 , \pm 60 , \pm 120$

Actual zeros:

$- \frac{3}{2}$, $8$, $\pm \sqrt{5}$

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{4} - 13 {x}^{3} - 34 {x}^{2} + 65 x + 120$

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 $120$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm \frac{5}{2} , \pm 3 , \pm 4 , \pm 5 , \pm 6 , \pm \frac{15}{2} , \pm 8 , \pm 10 , \pm 12 , \pm 15 , \pm 20 , \pm 24 , \pm 30 , \pm 40 , \pm 60 , \pm 120$

That's quite a few possibilities to try, so let's try a few easier ones to evaluate and deduce where the zeros might be...

$f \left(- 2\right) = 2 {\left(\textcolor{b l u e}{- 2}\right)}^{4} - 13 {\left(\textcolor{b l u e}{- 2}\right)}^{3} - 34 {\left(\textcolor{b l u e}{- 2}\right)}^{2} + 65 \left(\textcolor{b l u e}{- 2}\right) + 120$

$\textcolor{w h i t e}{f \left(- 2\right)} = 32 + 102 - 136 - 130 + 120 = - 12 < 0$

$f \left(- 1\right) = 2 + 13 - 34 - 65 + 120 = 36 > 0$

So $f \left(x\right)$ has a zero in $\left(- 2 , - 1\right)$. Is it the only possible rational one?

$f \left(- \frac{3}{2}\right) = 2 {\left(\textcolor{b l u e}{- \frac{3}{2}}\right)}^{4} - 13 {\left(\textcolor{b l u e}{- \frac{3}{2}}\right)}^{3} - 34 {\left(\textcolor{b l u e}{- \frac{3}{2}}\right)}^{2} + 65 \left(\textcolor{b l u e}{- \frac{3}{2}}\right) + 120$

$\textcolor{w h i t e}{f \left(- \frac{3}{2}\right)} = 2 \left(\frac{81}{16}\right) + 13 \left(\frac{27}{8}\right) - 34 \left(\frac{9}{4}\right) - 65 \left(\frac{3}{2}\right) + 120$

$\textcolor{w h i t e}{f \left(- \frac{3}{2}\right)} = \frac{1}{8} \left(81 + 351 - 612 - 780 + 960\right) = 0$

So $x = - \frac{3}{2}$ is a zero and $\left(2 x + 3\right)$ a factor:

$2 {x}^{4} - 13 {x}^{3} - 34 {x}^{2} + 65 x + 120 = \left(2 x + 3\right) \left({x}^{3} - 8 {x}^{2} - 5 x + 40\right)$

The remaining cubic can be factored by grouping:

${x}^{3} - 8 {x}^{2} - 5 x + 40 = \left({x}^{3} - 8 {x}^{2}\right) - \left(5 x - 40\right)$

$\textcolor{w h i t e}{{x}^{3} - 8 {x}^{2} - 5 x + 40} = {x}^{2} \left(x - 8\right) - 5 \left(x - 8\right)$

$\textcolor{w h i t e}{{x}^{3} - 8 {x}^{2} - 5 x + 40} = \left({x}^{2} - 5\right) \left(x - 8\right)$

$\textcolor{w h i t e}{{x}^{3} - 8 {x}^{2} - 5 x + 40} = \left({x}^{2} - {\left(\sqrt{5}\right)}^{2}\right) \left(x - 8\right)$

$\textcolor{w h i t e}{{x}^{3} - 8 {x}^{2} - 5 x + 40} = \left(x - \sqrt{5}\right) \left(x + \sqrt{5}\right) \left(x - 8\right)$

So the other three zeros are $\pm \sqrt{5}$ and $8$