# What are all the possible rational zeros for f(x)=8x^3-2x+24 and how do you find all zeros?

Aug 29, 2016

$f \left(x\right)$ has zeros:

${x}_{1} = - \frac{3}{2}$

${x}_{2 , 3} = \frac{3}{4} \pm \frac{\sqrt{23}}{4} i$

#### Explanation:

$f \left(x\right) = 8 {x}^{3} - 2 x + 24 = 2 \left(4 {x}^{3} - x + 12\right)$

By the rational roots theorem, any rational zeros of $4 {x}^{3} - x + 12$ 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 $4$ of the leading term.

So the only possible rational zeros of $4 {x}^{3} - x + 12$ and thus $f \left(x\right)$ are:

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

By Descartes' rule of signs we can tell that $f \left(x\right)$ has $0$ or $2$ positive zeros and $1$ negative zero.

So let's try the negative possibilities first:

$f \left(- \frac{1}{4}\right) = 2 \left(4 \left(- \frac{1}{64}\right) - \left(- \frac{1}{4}\right) + 12\right) = 2 \left(- \frac{1}{16} + \frac{1}{4} + 12\right) \ne 0$

$f \left(- \frac{1}{2}\right) = 2 \left(4 \left(- \frac{1}{8}\right) - \left(- \frac{1}{2}\right) + 12\right) = 2 \left(- \frac{1}{2} + \frac{1}{2} + 12\right) = 24 \ne 0$

$f \left(- \frac{3}{4}\right) = 2 \left(4 \left(- \frac{27}{64}\right) - \left(- \frac{3}{4}\right) + 12\right) = 2 \left(- \frac{27}{16} + \frac{3}{4} + 12\right) \ne 0$

$f \left(- 1\right) = 2 \left(- 4 + 1 + 12\right) = 18 \ne 0$

$f \left(- \frac{3}{2}\right) = 2 \left(4 \left(- \frac{27}{8}\right) - \left(- \frac{3}{2}\right) + 12\right) = 2 \left(- \frac{27}{2} + \frac{3}{2} + \frac{24}{2}\right) = 0$

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

$4 {x}^{3} - x + 12 = \left(2 x + 3\right) \left(2 {x}^{2} - 3 x + 4\right)$

$x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(2\right) \left(4\right)}}{2 \cdot 2}$
$= \frac{3 \pm \sqrt{9 - 32}}{4}$
$= \frac{3 \pm \sqrt{- 23}}{4}$
$= \frac{3 \pm \sqrt{23} i}{4}$
$= \frac{3}{4} \pm \frac{\sqrt{23}}{4} i$