College Algebra, Chapter 4, 4.4, Section 4.4, Problem 46

Determine all the real zeros of the polynomial $P(x) = x^3 - 5x^2 + 2x + 12$. Use the quadratic formula if necessary.

The leading coefficient of $P$ is $1$, so all the rational zeros are integers. They are the divisors of constant term $12$. Thus, the possible candidates are

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Using Synthetic Division,







We find that $1$ and $2$ are not zeros, but that $3$ is a zero and that $P$ factors as

$x^3 - 5x^2 + 2x + 12 = (x - 3)(x^2 - 2x - 4)$

We now factor the quotient $x^2 - 2x - 4$ using the quadratic formula


$
\begin{equation}
\begin{aligned}

x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\
\\
x =& \frac{-(-2) \pm \sqrt{(-2)^2 - 4 (1)(-4)}}{2(1)}
\\
\\
x =& 1 \pm \sqrt{5}

\end{aligned}
\end{equation}
$


The zeros of $P$ are $3, 1 + \sqrt{5}$ and $1 - \sqrt{5}$.