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

Oct 25, 2016

$f \left(x\right)$ has zeros $3$ and $- \frac{1}{6} \pm \frac{\sqrt{11}}{6} i$

#### Explanation:

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

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

That means that the only possible rational zeros are:

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

In addition note that the signs of the coefficients of $f \left(x\right)$ are in the pattern $+ - - -$. By Descartes' Rule of signs, since there is one change of sign, $f \left(x\right)$ has exactly one positive Real zero.

So check the possible positive rational zeros first:

$f \left(\frac{1}{3}\right) = 3 \left(\frac{1}{27}\right) - 8 \left(\frac{1}{9}\right) - 2 \left(\frac{1}{3}\right) - 3 = \frac{1 - 8 - 6 - 27}{9} = - \frac{40}{9}$

$f \left(1\right) = 3 - 8 - 2 - 3 = - 10$

$f \left(3\right) = 3 \left(27\right) - 8 \left(9\right) - 2 \left(3\right) - 3 = 81 - 72 - 6 - 3 = 0$

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

$3 {x}^{3} - 8 {x}^{2} - 2 x - 3 = \left(x - 3\right) \left(3 {x}^{2} + x + 1\right)$

Use the quadratic formula to solve $3 {x}^{2} + x + 1 = 0$ to find the two remaining zeros:

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

$\textcolor{w h i t e}{x} = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(3\right) \left(1\right)}}{2 \cdot 3}$

$\textcolor{w h i t e}{x} = \frac{- 1 \pm \sqrt{1 - 12}}{6}$

$\textcolor{w h i t e}{x} = \frac{- 1 \pm \sqrt{- 11}}{6}$

$\textcolor{w h i t e}{x} = - \frac{1}{6} \pm \frac{\sqrt{11}}{6} i$