Answer
The zeros are $x=-\frac{1}{2}$ with a multiplicity of 1 and $x=4$ with a multiplicity of 3. At $x=-\frac{1}{2}$ and $x=4$, the graph will cross the $x\text{-axis}$.
Work Step by Step
For zeros, let $f\left( x \right)=0$.
That is,
$\begin{align}
& -3\left( x+\frac{1}{2} \right){{\left( x-4 \right)}^{3}}=0 \\
& \left( x+\frac{1}{2} \right){{\left( x-4 \right)}^{3}}=0.
\end{align}$
Then, the values of x are as follows:
$x=-\frac{1}{2},4$.
For multiplicity:
$x=-\frac{1}{2}$, with a multiplicity of 1, because of $\left( x+\frac{1}{2} \right)$.
$x=4$, with a multiplicity of 3, because of ${{\left( x-4 \right)}^{3}}$.
The multiplicity of $x=-\frac{1}{2}$ is odd; the graph will cross only the $x\text{-axis}$.
The multiplicity of $x=4$ is odd; the graph will cross only the $x\text{-axis}$.
