College Algebra, Chapter 4, 4.5, Section 4.5, Problem 48

Determine all the zeros of the polynomial $P(x) = x^3 + 7x^2 + 18x + 18$.
The possible rational zeros are the factors of 18 which are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 9$ and $\pm 18$. By using synthetic division and by trial and error,


Thus,

$
\begin{equation}
\begin{aligned}
P(x) &= x^3 + 7x^2 + 18x + 18\\
\\
&= (x+3)(x^2+4x+6)
\end{aligned}
\end{equation}
$


To find the complex roots, we use quadratic formula.

$
\begin{equation}
\begin{aligned}
x &= \frac{-4 \pm \sqrt{4^2 - 4(1)(6)}}{2(1)}\\
\\
&= \frac{-4 \pm \sqrt{-8}}{2} = -2\pm\sqrt{2}i
\end{aligned}
\end{equation}
$

Thus, the zeros of $P$ are $-3, -2\pm\sqrt{2}i$ and $-2 -\sqrt{2}i$