Answer
The zeros are $x=3$ with multiplicity 1, $x=-3$ with multiplicity 1 and $x=-5$ with multiplicity 1. At $x=3$, $x=-3$, and $x=-5$, the graph will cross the $x\text{-axis}$.
Work Step by Step
For zeros, let $f\left( x \right)=0$.
That is,
$\begin{align}
& {{x}^{3}}+5{{x}^{2}}-9x-45=0 \\
& {{x}^{2}}\left( x+5 \right)-9\left( x+5 \right)=0 \\
& \left( {{x}^{2}}-9 \right)\left( x+5 \right)=0 \\
& \left( x-3 \right)\left( x+3 \right)\left( x+5 \right)=0.
\end{align}$
Then, the values of x are as follows:
$x=3,-3,-5.$
For multiplicity,
$x=3$ with a multiplicity 1, because of $\left( x-3 \right)$.
$x=-3$ with a multiplicity 1, because of $\left( x+3 \right)$.
$x=-5$ with a multiplicity 1, because of $\left( x+5 \right)$.
The multiplicity of $x=3$ is odd; the graph will cross the $x\text{-axis}$.
The multiplicity of $x=-3$ is odd; the graph will cross the $x\text{-axis}$.
The multiplicity of $x=-5$ is odd; the graph will cross the $x\text{-axis}$.
