Answer
The zeros are $x=2$ with multiplicity 1, $x=-2$ with multiplicity 1 and $x=-7$ with multiplicity 1. At $x=2$, $x=-2$, and $x=-7$, the graph will cross the $x\text{-axis}$.
Work Step by Step
For zeros, let $f\left( x \right)=0$.
That is,
$\begin{align}
& {{x}^{3}}+7{{x}^{2}}-4x-28=0 \\
& {{x}^{2}}\left( x+7 \right)-4\left( x+7 \right)=0 \\
& \left( {{x}^{2}}-4 \right)\left( x+7 \right)=0 \\
& \left( x+2 \right)\left( x-2 \right)\left( x+7 \right)=0.
\end{align}$
Then, the values of x are as follows:
$x=2,-2,-7$.
For multiplicity,
$x=2$, with a multiplicity 1, because of $\left( x-2 \right)$.
$x=-2$, with a multiplicity 1, because of $\left( x+2 \right)$.
$x=-7$, with a multiplicity 1, because of. $\left( x+7 \right)$.
The multiplicity of $x=2$ is odd; the graph will cross the $x\text{-axis}$.
The multiplicity of $x=-2$ is odd; the graph will cross the $x\text{-axis}$.
The multiplicity of $x=-7$ is odd; the graph will cross the $x\text{-axis}$.
