College Algebra, Chapter 4, 4.4, Section 4.4, Problem 56

The polynomial $P(x) = -x^3 - 2x^2 + 5x + 6$.

a.) Find all the real zeros of $P$

The leading coefficient of $P$ is $-1$, so all rational zeros are integer: they are divisors of constant term $6$. Thus, the possible zeros are

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

Using Synthetic Division







We find that $1$ is not a zero but that $2$ is a zero and that $P$
factors as

$\displaystyle -x^3 - 2x^2 + 5x + 6 = (x - 2) (-x^2 - 4x - 3) $

We now factor the quotient $-x^2 - 4x - 3$ using quadratic
formula


$
\begin{equation}
\begin{aligned}

x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\
\\
x =& \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(-3)}}{2(-1)}
\\
\\
x =& -1 \text{ and } x = -3

\end{aligned}
\end{equation}
$


The zeros of $P$ are $2, -1$ and $-3$.

b.) Sketch the graph of $P$