#### Answer

The zeros of $f\left( x \right)=2{{x}^{3}}-2x$ are $0,1,-1$.

#### Work Step by Step

First, equate the given function to $0$ as $f\left( x \right)=0$. So,
$2{{x}^{3}}-2x=0$
Then, the function $f\left( x \right)=2{{x}^{3}}-2x$ can be factorized as:
$2x\left( {{x}^{2}}-1 \right)=0$
By using the formula, $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ the polynomial ${{x}^{2}}-1$ is written as,
${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$
So, the provided function factorizes as:
$2x\left( x+1 \right)\left( x-1 \right)=0$
Equating each factor to $0$. So,
$\begin{align}
& 2x=0 \\
& x=0
\end{align}$
Or,
$\left( x-1 \right)=0$
Or,
$\left( x+1 \right)=0$
Thus the zeros of the given equation are,
$x=0,1,-1$
Also, the given function crosses the x-axis at $x=0,1,-1$ as it has a multiplicity of 1.
Since the function is an odd-degree function and the leading coefficient is 2, so the graph falls to the left and rises to the right.