College Algebra, Chapter 4, 4.4, Section 4.4, Problem 58

The polynomial $P(x) = 3x^3 + 17x^2 + 21x - 9$.

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

The leading coefficient of $P$ is $3$, and its factors are $\pm 1, \pm 3$. They are divisors of constant term $-9$. Its factors are $\pm 1, \pm 3, \pm 9$. Thus, the possible zeros are

$\displaystyle \pm 1, \pm 3, \pm \frac{1}{3}, \pm \frac{1}{9}$

Using Synthetic Division







We find that $1$ and $3$ are not zeros but that $\displaystyle \frac{1}{3}$ is zero and that $P$ factors as


$
\begin{equation}
\begin{aligned}

3x^3 + 17x^2 + 21x - 9 =& \left( x - \frac{1}{3} \right) (3x^2 + 18x + 27)
\\
\\
3x^3 + 17x^2 + 21x - 9 =& 3 \left( x - \frac{1}{3} \right) (x^2 + 6x + 9)

\end{aligned}
\end{equation}
$


We now factor the quotient $x^2 + 6x + 9$ using Sum of a Square, we get $\displaystyle 3x^3 + 17x^2 + 21x - 9 = 3 \left( x - \frac{1}{3} \right) (x + 3)^2$

The zeros of $P$ are $\displaystyle \frac{1}{3}$ and $-3$


b.) Sketch the graph of $P$