College Algebra, Chapter 4, 4.4, Section 4.4, Problem 34

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

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

Using Synthetic Division







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

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

We now factor the quotient $2x^2 - x - 3$ using trial and error. We get,


$
\begin{equation}
\begin{aligned}

2x^3 - 3x^2 - 2x + 3 =& (x - 1) (2x - 3) (x + 1)

\end{aligned}
\end{equation}
$


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