A rational expression is a fraction whose top and bottom are polynomials, like $\frac{x^2 - 9}{x^2 + x - 6}$. Simplifying one works exactly like reducing $\frac{12}{18}$ to $\frac{2}{3}$: factor, then cancel common factors. You will use everything from Factoring Completely.
Key idea
Factor the numerator and denominator completely, then cancel factors that appear in both. Before you cancel, note the excluded values: every $x$ that makes the original denominator 0. The simplified expression equals the original only for $x$ outside that list.
Why it works
Cancelling uses the fact that $\frac{ac}{bc} = \frac{a}{b} \cdot \frac{c}{c} = \frac{a}{b}$, because $\frac{c}{c} = 1$. That is only true when $c \ne 0$. So when you cancel $(x - 3)$, you are quietly assuming $x \ne 3$. The original expression is undefined at $x = 3$ and stays undefined there; the graph shows a hole at that point.
Worked examples
Worked example: Factor, then cancel
Simplify $\frac{x^2 - 9}{x^2 + x - 6}$ and state the excluded values.
- Factor: $\frac{(x - 3)(x + 3)}{(x + 3)(x - 2)}$.
- Excluded values come from the original denominator: $x \ne -3$ and $x \ne 2$.
- Cancel $(x + 3)$.
Answer: $\frac{x - 3}{x - 2}$, for $x \ne -3, 2$
Worked example: Opposite factors
Simplify $\frac{5 - x}{x^2 - 25}$.
- Factor: $x^2 - 25 = (x - 5)(x + 5)$, and $5 - x = -(x - 5)$.
- Cancel $(x - 5)$, keeping the $-1$: $\frac{-1}{x + 5}$.
Answer: $-\frac{1}{x + 5}$, for $x \ne \pm 5$
Common mistake
Cancelling terms instead of factors, as in $\frac{x + 3}{3} = x$. The 3 on top is added, not multiplied, so it can't cancel. Test with a number: at $x = 3$, $\frac{3 + 3}{3} = 2$, not 3. Only cancel something that multiplies the entire numerator and the entire denominator.