## Precalculus (6th Edition) Blitzer

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