# How do you use the rational roots theorem to find all possible zeros of f(x)=2x^3-7x^2-10x-3?

Aug 8, 2016

$f \left(x\right)$ has zeros $- \frac{1}{2}$ and $2 \pm \sqrt{7}$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 7 {x}^{2} - 10 x - 3$

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 $- 3$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

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

We find:

$f \left(- \frac{1}{2}\right) = 2 {\left(- \frac{1}{2}\right)}^{3} - 7 {\left(- \frac{1}{2}\right)}^{2} - 10 \left(- \frac{1}{2}\right) - 3$

$= - \frac{1}{4} - \frac{7}{4} + \frac{10}{2} - 3 = 0$

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

$2 {x}^{3} - 7 {x}^{2} - 10 x - 3$

$= \left(2 x + 1\right) \left({x}^{2} - 4 x - 3\right)$

$= \left(2 x + 1\right) \left({x}^{2} - 4 x + 4 - 7\right)$

$= \left(2 x + 1\right) \left({\left(x - 2\right)}^{2} - {\left(\sqrt{7}\right)}^{2}\right)$

$= \left(2 x + 1\right) \left(x - 2 - \sqrt{7}\right) \left(x - 2 + \sqrt{7}\right)$

Hence the two other zeros are:

$x = 2 \pm \sqrt{7}$