A rational expression looks fancy, but it is hiding nothing new. It is just a fraction whose top and bottom happen to be polynomials — and it plays by the very same rules you already trust for ordinary fractions.
Start with a fraction you already know
Before we touch a single \( x \), let's remember how an ordinary fraction behaves. A rational expression obeys the very same rules — only now the top and bottom are polynomials.
Take \( \frac{3}{4} \). It means three of four equal pieces. You can shade it, cut its pieces finer, or reduce it to lowest terms, and nothing about its value changes. Drag the slider below and watch the shaded amount stay put while the pieces rearrange.
Now compare these two objects side by side:
\[ \underbrace{\frac{3}{4}}_{\text{numeric fraction}} \qquad\qquad \underbrace{\frac{x^2 - 9}{x + 3}}_{\text{rational expression}} \]The one on the left has plain numbers on top and bottom. The one on the right has polynomials — expressions built from \( x \) — on top and bottom. That is the only difference. A rational expression is a ratio of two polynomials, exactly the way a rational number is a ratio of two integers. Everything you know about \( \frac{3}{4} \) is about to carry over.
The word "rational" means ratio. A rational number is a ratio of integers, like \( \frac{3}{4} \). A rational expression is a ratio of polynomials, like \( \frac{x^2-9}{x+3} \). Same idea, richer ingredients.
Simplifying: factor, then cancel
How did you reduce \( \frac{6}{8} \) to \( \frac{3}{4} \)? You found the factor both numbers shared — a \( 2 \) — and cancelled it:
\[ \frac{6}{8} = \frac{2 \times 3}{2 \times 4} = \frac{3}{4} \]Rational expressions reduce the very same way, with one extra step at the front: you have to factor the polynomials first so the shared piece becomes visible. Once both top and bottom are written as products, any factor appearing in both cancels:
\[ \frac{x^2 - 9}{x + 3} = \frac{(x-3)(x+3)}{x+3} = x - 3 \]The \( (x+3) \) on top and the \( (x+3) \) on the bottom are the same quantity, so their ratio is \( 1 \) and they vanish — exactly like the shared \( 2 \) in \( \frac{6}{8} \).
The forbidden values
Here is the one genuinely new idea, and it is the heart of the lesson. A denominator can never be zero — division by zero is undefined. So before you cancel anything, look at the original bottom and ask: which values of \( x \) would make it zero? Those values are forbidden, and they stay forbidden even after the offending factor is cancelled away.
In \( \dfrac{x^2-9}{x+3} \), the original denominator \( x+3 \) is zero when \( x = -3 \). That value was never allowed. After cancelling we are left with the tidy \( x - 3 \), which looks perfectly happy at \( x = -3 \) — but it is not. The honest, complete answer keeps the restriction in plain sight:
\[ \frac{x^2 - 9}{x + 3} = x - 3, \qquad x \neq -3 \]Think of it as a tiny pinhole in the graph: the simplified line \( y = x - 3 \) is correct everywhere except at \( x = -3 \), where the original expression simply has no value. We write \( x \neq -3 \) to remember the hole is there.
Tip Find the forbidden values from the original denominator, before you cancel. Set the bottom equal to zero, solve, and exclude those \( x \)-values — they ride along with your simplified answer for the rest of its life.
The four operations carry over
Every rule you learned for numeric fractions transfers without change. Only the "find the forbidden values" habit comes along for the ride.
Multiply straight across. Numerators times numerators, denominators times denominators — and cancel common factors to keep things tidy:
\[ \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \]Divide by flipping. Dividing by a fraction is the same as multiplying by its reciprocal — flip the second one and multiply:
\[ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} \]Add and subtract only over a common denominator. Just as you cannot add halves to thirds until both are sixths, you cannot combine \( \frac{1}{x} \) and \( \frac{1}{x+1} \) until they share a bottom. Rewrite both over the common denominator \( x(x+1) \), then add the tops:
\[ \frac{1}{x} + \frac{1}{x+1} = \frac{x+1}{x(x+1)} + \frac{x}{x(x+1)} = \frac{2x+1}{x(x+1)} \]Same rule, same picture — only the pieces are now made of \( x \).
Meet the vertical asymptote
What does a forbidden value look like? When the denominator approaches zero, the fraction's value blows up — and the graph develops a vertical asymptote, a line the curve races toward but never reaches.
The widget below plots \( y = \dfrac{a}{x - h} + k \). Right now it shows \( \dfrac{1}{x - 2} \): the denominator is zero at \( x = 2 \), so that is the forbidden value. Watch what happens near the dashed vertical line at \( x = 2 \) — as \( x \) creeps toward \( 2 \), the denominator shrinks toward zero and the value shoots off to \( +\infty \) on one side and \( -\infty \) on the other. The curve hugs the line ever closer but never touches it. Slide \( h \) and \( k \) and notice the dashed lines move with the asymptotes.
The number \( h \) slides the forbidden value (and its vertical asymptote) left and right; the number \( k \) lifts the whole curve up and down, carrying the horizontal asymptote with it. The story is always the same: wherever the denominator hits zero, the graph cannot follow — so it flees to infinity instead.
Asymptotes are forbidden values made visible. A vertical asymptote sits exactly at an \( x \)-value the denominator forbids. The algebra (\( x \neq 2 \)) and the geometry (the dashed line at \( x = 2 \)) are two views of the very same fact.
Worked examples
- Factor the top. The numerator \( x^2 - 9 \) is a difference of squares: \( x^2 - 9 = (x-3)(x+3) \).
- Note the forbidden value before cancelling: the original denominator \( x+3 \) is zero at \( x = -3 \), so \( x \neq -3 \).
- Cancel the shared factor \( (x+3) \) from top and bottom: \( \dfrac{(x-3)(x+3)}{x+3} = x - 3 \).
The simplified result is \( \mathbf{x - 3}, \ x \neq -3 \). The restriction is part of the answer — without it the statement would be false at \( x = -3 \).
- List the forbidden values from the original denominators: \( x+1 = 0 \) gives \( x \neq -1 \), and \( x^2 = 0 \) gives \( x \neq 0 \).
- Multiply straight across: \( \dfrac{x \cdot (x+1)}{(x+1) \cdot x^2} = \dfrac{x(x+1)}{x^2(x+1)} \).
- Cancel the shared \( (x+1) \), and cancel one \( x \) from top against \( x^2 \) on the bottom: \( \dfrac{x(x+1)}{x^2(x+1)} = \dfrac{1}{x} \).
The product is \( \mathbf{\dfrac{1}{x}}, \ x \neq 0,\, -1 \). Notice \( -1 \) survives in the restriction even though no \( (x+1) \) remains in sight — it was forbidden from the start.
Practice
Try each one yourself, then reveal the full solution.
1. Simplify \( \dfrac{x^2 - 1}{x - 1} \).
Factor the top as a difference of squares: \( x^2 - 1 = (x-1)(x+1) \).
Read the forbidden value off the original denominator: \( x - 1 = 0 \) gives \( x \neq 1 \).
Cancel the shared \( (x-1) \): \( \dfrac{(x-1)(x+1)}{x-1} = x + 1 \).
So \( \dfrac{x^2 - 1}{x - 1} = \mathbf{x + 1}, \ x \neq 1 \).
2. Multiply \( \dfrac{2x}{3} \cdot \dfrac{9}{4x} \).
The denominator \( 4x \) is zero at \( x = 0 \), so \( x \neq 0 \).
Multiply straight across: \( \dfrac{2x \cdot 9}{3 \cdot 4x} = \dfrac{18x}{12x} \).
Cancel the shared \( x \), then reduce \( \dfrac{18}{12} \) by its greatest common factor \( 6 \): \( \dfrac{18}{12} = \dfrac{3}{2} \).
So \( \dfrac{2x}{3} \cdot \dfrac{9}{4x} = \mathbf{\dfrac{3}{2}}, \ x \neq 0 \).
3. Add \( \dfrac{1}{x} + \dfrac{1}{x+1} \).
The pieces are different sizes, so build a common denominator: \( x(x+1) \). (Forbidden values: \( x \neq 0 \) and \( x \neq -1 \).)
Rewrite each fraction over that bottom: \( \dfrac{1}{x} = \dfrac{x+1}{x(x+1)} \) and \( \dfrac{1}{x+1} = \dfrac{x}{x(x+1)} \).
Now the bottoms match, so add the tops: \( \dfrac{(x+1) + x}{x(x+1)} = \dfrac{2x+1}{x(x+1)} \).
So \( \dfrac{1}{x} + \dfrac{1}{x+1} = \mathbf{\dfrac{2x+1}{x(x+1)}} \).