## Precalculus (6th Edition) Blitzer

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