#### Answer

The zeros of $f\left( x \right)={{x}^{3}}-{{x}^{2}}-4x+4$ are $1,2,-2$.

#### Work Step by Step

First put $f\left( x \right)=0$. So,
Consider, ${{x}^{3}}-{{x}^{2}}-4x+4$
Factorize it,
$\begin{align}
& {{x}^{3}}-{{x}^{2}}-4x+4=0 \\
& {{x}^{2}}\left( x-1 \right)-4\left( x-1 \right)=0 \\
& \left( {{x}^{2}}-4 \right)\left( x-1 \right)=0
\end{align}$
By using the formula $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ the polynomial ${{x}^{2}}-4$ is written as:
${{x}^{2}}-4=\left( x-2 \right)\left( x+2 \right)$
So, the provided function factorizes as:
$\left( x-1 \right)\left( x-2 \right)\left( x+2 \right)=0$
Put each factor equal to $0$. So,
$\left( x-1 \right)=0$
Or,
$\left( x-2 \right)=0$
Or,
$\left( x+2 \right)=0$
The above equations provide the values of x as,
$x=1,2,-2$
The graph of the function passes the x-axis at $x=1,2,-2$ since all the roots have a multiplicity of 1.
Since it is an odd degree polynomial (degree=3) and the leading coefficient is 1, the graph falls to the left and rises to the right.