College Algebra, Chapter 4, 4.4, Section 4.4, Problem 38

Determine all rational zeros of the polynomial $P(x) = 6x^3 + 11x^2 - 3x - 2$, and write the polynomial in factored form.

The leading coefficient of $P$ is $6$ and its factors are $\pm 1, \pm 2, \pm 3, \pm 6$. They are the divisors of the constant term $-2$ and its factors are $\pm 1, \pm 2$. The possible rational zeros are $\displaystyle \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6}$

Using Synthetic Division







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


$
\begin{equation}
\begin{aligned}

6x^3 + 11x^2 - 3x - 2 =& \left( x - \frac{1}{2} \right) (6x^2 + 14x + 4)
\\
\\
6x^3 + 11x^2 - 3x - 2 =& 2 \left( x - \frac{1}{2} \right) (3x^2 + 7x + 2)

\end{aligned}
\end{equation}
$


We now factor $3x^2 + 7x + 2$ using trial and error. We get,


$
\begin{equation}
\begin{aligned}

6x^3 + 11x^2 - 3x - 2 =& 2 \left( x - \frac{1}{2} \right) (3x + 1)(x + 2)

\end{aligned}
\end{equation}
$


The zeros of $P$ are $\displaystyle \frac{1}{2}, \frac{-1}{3}$ and $-2$.