## Precalculus (6th Edition) Blitzer

The domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ is $\left( -\infty ,-2 \right)\cup \left( -2,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.
In the given function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ if the denominator is zero then it will make the function undefined. So, put the denominator ${{x}^{3}}-5{{x}^{2}}-4x+20$ equal to zero. Therefore, ${{x}^{3}}-5{{x}^{2}}-4x+20=0$ Factorize the equation ${{x}^{3}}-5{{x}^{2}}-4x+20=0$ \begin{align} {{x}^{2}}\left( x-5 \right)-4\left( x-5 \right)=0 & \\ \left( {{x}^{2}}-4 \right)\left( x-5 \right)=0 & \\ \end{align} Case 1: \begin{align} & {{x}^{2}}-4=0\ \ \ \ \ \ \ \ \\ & \left( x-2 \right)\left( x+\text{2} \right)\text{=}\ \text{0} \end{align} This implies $x=2$ and $x=-2$. Case 2: \begin{align} x-5=0 & \\ x=5 & \\ \end{align} The factors of the equation are $2\ ,-2$ and $5$. Exclude the factors from the domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$. Hence, the domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ is $\left( -\infty ,-2 \right)\cup \left( -2,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.