# What are all the possible rational zeros for f(x)=2x^3-8x^2+15x-27 and how do you find all zeros?

Sep 12, 2017

Possible rational zeros:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3 , \pm \frac{9}{2} , \pm 9 , \pm \frac{27}{2} , \pm 27$

Actual zeros:

$x = 3 \text{ }$ and $\text{ } x = \frac{1}{2} \left(1 \pm \sqrt{17} i\right)$

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{3} - 8 {x}^{2} + 15 x - 27$

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 $- 27$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3 , \pm \frac{9}{2} , \pm 9 , \pm \frac{27}{2} , \pm 27$

Using Descartes' Rule of Signs, note that the pattern of the signs of the coefficients of $f \left(x\right)$ is $+ - + -$. With $3$ changes of sign, we can tell that $f \left(x\right)$ has $3$ or $1$ positive real zeros. The pattern of the signs of the coefficients of $f \left(- x\right)$ is $- - - -$. With no changes, that means that $f \left(x\right)$ has no negative real zeros.

After a bit of trial and error, we find:

$f \left(3\right) = 2 {\left(\textcolor{b l u e}{3}\right)}^{3} - 8 {\left(\textcolor{b l u e}{3}\right)}^{2} + 15 \left(\textcolor{b l u e}{3}\right) - 27$

$\textcolor{w h i t e}{f \left(3\right)} = 54 - 72 + 45 - 27 = 0$

So $x = 3$ is a zero and $\left(x - 3\right)$ a factor:

$2 {x}^{3} - 8 {x}^{2} + 15 x - 27 = \left(x - 3\right) \left(2 {x}^{2} - 2 x + 9\right)$

Note that the discriminant of the remaining quadratic factor is negative, so it has non-real complex zeros.

We can complete the square (or use the quadratic formula) to find them.

For example:

$0 = 2 \left(2 {x}^{2} - 2 x + 9\right)$

$\textcolor{w h i t e}{0} = 4 {x}^{2} - 4 x + 1 + 17$

$\textcolor{w h i t e}{0} = {\left(2 x - 1\right)}^{2} + {\left(\sqrt{17}\right)}^{2}$

$\textcolor{w h i t e}{0} = {\left(2 x - 1\right)}^{2} - {\left(\sqrt{17} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 x - 1\right) - \sqrt{17} i\right) \left(\left(2 x - 1\right) + \sqrt{17} i\right)$

$\textcolor{w h i t e}{0} = \left(2 x - 1 - \sqrt{17} i\right) \left(2 x - 1 + \sqrt{17} i\right)$

So:

$x = \frac{1}{2} \left(1 \pm \sqrt{17} i\right)$