#### Answer

$f\left( x \right)={{\left( x+2 \right)}^{2}}\left( x-1 \right)$

#### Work Step by Step

It can be determined from the graph that the roots of the function are $-2,1$.
But, since the function is cubic, the function should have 3 roots.
Thus, one of the above roots will have multiplicity 2.
Now, since the graph is turning around at $x=-2$, this root will have multiplicity 2.
Then, the provided function can be factorized as follows:
$\begin{align}
& f\left( x \right)={{x}^{3}}+3{{x}^{2}}-4 \\
& f\left( x \right)={{\left( x+2 \right)}^{2}}\left( x-1 \right) \\
\end{align}$
Therefore, the function $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-4$ can be factorized as $f\left( x \right)={{\left( x+2 \right)}^{2}}\left( x-1 \right)$.