College Algebra, Chapter 4, 4.4, Section 4.4, Problem 52

Determine all the real zeros of the polynomial $P(x) = 3x^3 - 5x^2 - 8x - 2$. Use the quadratic formula if necessary.

The leading coefficient of $P$ is $3$, and its factors are $\pm 1, \pm 3$. They are the divisors of constant term $-2$ and its factors are $\pm 1, \pm 2$. Thus, the possible zeros are

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

Using Synthetic Division,







We find that $\displaystyle 1, 2, \frac{1}{3}, \frac{2}{3}, -1$ and $-2$ are not zeros but that $\displaystyle \frac{-1}{3}$ is a zero and that $P$ factors as

$\displaystyle 3x^3 - 5x^2 - 8x - 2 = \left( x + \frac{1}{3} \right) (3x^2 - 6x - 6)$

We now factor the quotient $3x^2 - 6x - 6$ using the quadratic formula


$\begin{equation} \begin{aligned} x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \\ x =& \frac{- (-6) \pm \sqrt{(-6)^2 - 4(3)(-6) }}{2(3)} \\ \\ x =& 1 \pm \sqrt{3} \end{aligned} \end{equation}$


The zeros of $P$ are $\displaystyle \frac{-1}{3}, 1 + \sqrt{3}$ and $1 - \sqrt{3}$.