# What are all the possible rational zeros for f(x)=10x^3-15x^2-16x+12 and how do you find all zeros?

Aug 7, 2017

The "possible" rational zeros are:

$\pm \frac{1}{10} , \pm \frac{1}{5} , \pm \frac{3}{10} , \pm \frac{2}{5} , \pm \frac{1}{2} , \pm \frac{3}{5} , \pm \frac{4}{5} , \pm 1 , \pm \frac{6}{5} , \pm \frac{3}{2} , \pm 2 , \pm \frac{12}{5} , \pm 3 , \pm 4 , \pm 6 , \pm 12$

The actual zeros are:

$2 \text{ }$ and $\text{ } - \frac{1}{4} \pm \frac{\sqrt{265}}{20}$

#### Explanation:

Given:

$f \left(x\right) = 10 {x}^{3} - 15 {x}^{2} - 16 x + 12$

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 $12$ and $q$ a divisor of the coefficient $10$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{10} , \pm \frac{1}{5} , \pm \frac{3}{10} , \pm \frac{2}{5} , \pm \frac{1}{2} , \pm \frac{3}{5} , \pm \frac{4}{5} , \pm 1 , \pm \frac{6}{5} , \pm \frac{3}{2} , \pm 2 , \pm \frac{12}{5} , \pm 3 , \pm 4 , \pm 6 , \pm 12$

Trying a few of the simpler values, we eventually find:

$f \left(2\right) = 10 {\left(\textcolor{b l u e}{2}\right)}^{3} - 15 {\left(\textcolor{b l u e}{2}\right)}^{2} - 16 \left(\textcolor{b l u e}{2}\right) + 12$

$\textcolor{w h i t e}{f \left(2\right)} = 80 - 60 - 32 + 12$

$\textcolor{w h i t e}{f \left(2\right)} = 0$

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

$10 {x}^{3} - 15 {x}^{2} - 16 x + 12 = \left(x - 2\right) \left(10 {x}^{2} + 5 x - 6\right)$

The remaining quadratic factor is in the standard form $a {x}^{2} + b x + c$ with $a = 10$, $b = 5$ and $c = - 6$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {5}^{2} - 4 \left(10\right) \left(- 6\right) = 25 + 240 = 265$

This is positive but not a perfect square. Hence we can deduce that the remaining zeros are real but irrational. We can find the zeros using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 5 \pm \sqrt{265}}{2 \cdot 10}$

$\textcolor{w h i t e}{x} = \frac{- 5 \pm \sqrt{265}}{20}$

$\textcolor{w h i t e}{x} = - \frac{1}{4} \pm \frac{\sqrt{265}}{20}$