Answer
The zeros are $x=-5$ with multiplicity 1 and $x=-2$ with a multiplicity 2. At $x=-5$ the graph will cross the $x\text{-axis}$ and at $x=-2$, the graph will touch the $x\text{-axis}$ and turn around.
Work Step by Step
For zeros:
Let $f\left( x \right)=0$.
That is,
$\begin{align}
& 3\left( x+5 \right){{\left( x+2 \right)}^{2}}\,=0 \\
& \left( x+5 \right){{\left( x+2 \right)}^{2}}=0.
\end{align}$
Then, for the values of x, we get,
$x=-5,-2$
Now, for multiplicity,
$x=-5$, with a multiplicity of 1, because of $\left( x+5 \right)$.
$x=-2$, with a multiplicity of 2, because of ${{\left( x+2 \right)}^{2}}$.
Since the multiplicity of $x=5$ is odd, the graph will cross only the $x\text{-axis}$.
And the multiplicity of $x=-4$ is even, so, the graph will touch the $x\text{-axis}$ and turn around.
