If [math]r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}[/math] is a rational function with a zero at [math]x=3[/math], what can you conclude?
[math]p\left(3\right)=0[/math]
The converse, however, is not true. That is, if [math]r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}[/math] is a rational function with [math]p\left(3\right)=0[/math], it is not necessarily the case that [math]r[/math] has a zero at [math]x=3[/math]. Plot an example of such a rational function below.
Plot a rational function [math]r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}[/math] below such that [math]p\left(-1\right)=p\left(2\right)=0[/math], and [math]r[/math]'s only zero occurs at [math]x=2[/math].
[math]x=a[/math] is a zero of the rational function [math]r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}[/math] if and only if:
[math]p\left(a\right)=0[/math], but [math]q\left(a\right)\ne0[/math].