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

If $P(x) = 9x^5 - 21x^4 + 10x^3 + 6x^2 - 3x - 1$, then

a.) Find all zeros of $P$, and state their multiplicities.

b.) Sketch the graph of $P$.



a.) To find the zeros of $P$, we apply synthetic division with the possible rational zeros of the factor of $1$ divided by the factor of $9$ which are $\displaystyle \pm 1, \pm \frac{1}{3}, \pm \frac{1}{9}$. Then, by trial and error







Again, by applying Synthetic Division







Again,







Again,








Thus,


$
\begin{equation}
\begin{aligned}

P(x) =& 9x^5 - 21x^4 + 10x^3 + 6x^2 - 3x - 1
\\
\\
=& \left(x + \frac{1}{3} \right) \left( 9x^4 - 24x^3 + 18x^2 - 3 \right)
\\
\\
=& \left( x + \frac{1}{3} \right) (x -1) \left(9x^3 - 15x^2 + 3x^2 + 3 \right)
\\
\\
=& \left( x + \frac{1}{3} \right) (x - 1)(x - 1) (9x^2 - 6x - 3)
\\
\\
=& \left( x + \frac{1}{3} \right) (x - 1)(x - 1)(x - 1) (9x + 3)
\\
\\
=& \left( x + \frac{1}{3} \right) (x - 1)^3 (9x + 3)

\end{aligned}
\end{equation}
$



Therefore, rational zeros of $P$ are $\displaystyle \frac{-1}{3}$ and $1$. Then, the zeros have multiplicity of $1$ and $3$ respectively.

b.) To sketch the graph of $P$, we must know first the intercepts of the function. The values of the $x$ intercepts are the zeros of the function, that is $\displaystyle \frac{-1}{3}$ and $1$. Next, to determine the $y$ intercept, we set $x = 0$ so that


$
\begin{equation}
\begin{aligned}

P(0) =& \left( 0 + \frac{1}{3} \right) (0 - 1)^3 (9(0) + 3)
\\
\\
=& \left( \frac{1}{3} \right) (-1)^3 (3)
\\
\\
=& -1

\end{aligned}
\end{equation}
$



Since the function has an odd degree and a positive leading coefficient, then its end behavior is $y \to - \infty$ as $x \to -\infty$ and $y \to \infty$ as $x \to \infty$. Then, the graph is