College Algebra, Chapter 4, 4.4, Section 4.4, Problem 32

Determine all rational zeros of the polynomial $P(x) = 2x^3 + 7x^2 + 4x - 4$, 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 $-4$ and the factors of $-4$ are $\pm 1, \pm 2, \pm 4$. The possible rational zeros are $\displaystyle \pm 1, \pm 2, \pm 4, \pm \frac{1}{2}$

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

$2x^3 + 7x^2 + 4x - 4 = \left( x - \frac{1}{2} \right) \left( 2x^2 + 8x + 8 \right)$

We now factor the quotient $2x^2 + 8x + 8$ using trial and error. We get,


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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