College Algebra, Chapter 4, 4.5, Section 4.5, Problem 46

Determine all the zeros of the polynomial $P(x) = x^3 - 7x^2 + 17x - 15$.
The possible rational zeros are the factors of 15 which are $\pm 1, \pm 3, \pm 5$ and $\pm 15$. By using synthetic division and by trial and error,


Thus,

$
\begin{equation}
\begin{aligned}
P(x) &= x^3 - 7x^2 + 17x - 15\\
\\
&= (x-3)(x^2 - 4x + 5)
\end{aligned}
\end{equation}
$


To find the complex zeros of $P$, we use quadratic formula.

$
\begin{equation}
\begin{aligned}
x &= \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}\\
\\
&= \frac{4 \pm \sqrt{-4}}{2} = \frac{4\pm 2\sqrt{-1}}{2} = 2 \pm i
\end{aligned}
\end{equation}
$

Thus, the zeros of $P$ are $3,2 +i$ and $2 - i$