## Precalculus (6th Edition) Blitzer

The zeros are $x=2$ with multiplicity 1, $x=-2$ with multiplicity 1 and $x=-7$ with multiplicity 1. At $x=2$, $x=-2$, and $x=-7$, the graph will cross the $x\text{-axis}$.
For zeros, let $f\left( x \right)=0$. That is, \begin{align} & {{x}^{3}}+7{{x}^{2}}-4x-28=0 \\ & {{x}^{2}}\left( x+7 \right)-4\left( x+7 \right)=0 \\ & \left( {{x}^{2}}-4 \right)\left( x+7 \right)=0 \\ & \left( x+2 \right)\left( x-2 \right)\left( x+7 \right)=0. \end{align} Then, the values of x are as follows: $x=2,-2,-7$. For multiplicity, $x=2$, with a multiplicity 1, because of $\left( x-2 \right)$. $x=-2$, with a multiplicity 1, because of $\left( x+2 \right)$. $x=-7$, with a multiplicity 1, because of. $\left( x+7 \right)$. The multiplicity of $x=2$ is odd; the graph will cross the $x\text{-axis}$. The multiplicity of $x=-2$ is odd; the graph will cross the $x\text{-axis}$. The multiplicity of $x=-7$ is odd; the graph will cross the $x\text{-axis}$.