## Precalculus (6th Edition) Blitzer

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