Rational Expressions

Simplifying a rational expression is a fixed routine: find the restrictions, factor, cancel only common factors, then carry the restrictions into your answer. Skip the first step and you can simplify correctly yet still write a wrong domain.

A rational expression is a fraction whose numerator and denominator are polynomials, such as $\frac{x+1}{x-3}$. In general it has the form

$$\frac{P(x)}{Q(x)}$$

where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \ne 0$. Because the denominator cannot be zero, every rational expression comes with values that are not allowed.

When this method applies

Use this simplification routine whenever you have a quotient of polynomials. Expressions such as

$$\frac{x+2}{x-5}, \quad \frac{x^2-1}{x^2+x}, \quad \frac{3}{x^2+4}$$

all qualify, because each is a polynomial over a polynomial. By contrast, $\frac{1}{\sqrt{x}}$ is not usually treated as a rational expression in basic algebra, since $\sqrt{x}$ is not a polynomial, so the cancel-factors routine does not apply the same way.

The procedure, step by step

1. Find restrictions first. Determine which values make the original denominator zero. Those values are excluded no matter how simple the expression later looks.
2. Factor before canceling. Rewrite numerator and denominator as products so a common factor becomes visible.
3. Cancel factors, not terms. Divide out only matching factors. You cannot cancel part of a sum or difference.
4. Keep the restrictions. Carry every original restriction into the final answer.

Step 3 is the heart of the rule. For example,

$$\frac{x+1}{x+3}$$

does not simplify by "canceling the $x$." The numerator and denominator are sums, not matching factors. That is why factoring comes first: factoring shows whether a common factor actually exists.

A full worked example

Simplify

$$\frac{x^2-1}{x^2+x}.$$

Step 1, restrictions. Set the original denominator to zero:

$$x^2 + x = x(x+1),$$

so $x \ne 0$ and $x \ne -1$.

Step 2, factor both parts.

$$x^2-1 = (x-1)(x+1), \qquad x^2+x = x(x+1).$$

So the expression becomes

$$\frac{(x-1)(x+1)}{x(x+1)}.$$

Step 3, cancel the common factor $(x+1)$.

$$\frac{x-1}{x}.$$

Step 4, keep the restrictions. The simplified expression is $\frac{x-1}{x}$, with the original restrictions $x \ne 0$ and $x \ne -1$.

Where the procedure breaks down

The factor $(x+1)$ disappeared from the final fraction, but the restriction $x \ne -1$ did not. This is the step students miss most. The original expression was undefined at $x=-1$, so the simplified answer must still exclude it.

Why this matters: the restriction changes the expression's domain, meaning the set of inputs that make sense. The form

$$\frac{x-1}{x}$$

looks defined at $x=-1$, but when it comes from $\frac{x^2-1}{x^2+x}$, that value stays excluded. Simplification changes appearance, not the points where the original was undefined.

Other common slips, by step:

1. Step 3: canceling terms instead of factors — the most common algebra error here.
2. Step 2: skipping the factoring, which hides whether cancellation is even legal.
3. Step 4: dropping the denominator restrictions after simplifying.
4. Assuming every rational expression simplifies. Some are already in simplest form.

Where this shows up

Rational expressions appear in algebra, precalculus, and calculus: simplifying formulas, solving rational equations, locating vertical asymptotes, and setting up partial fraction decomposition. Many formulas are ratios, so once factoring, simplifying, and restriction-tracking are automatic, later topics get much easier.

Run the steps yourself

Simplify

$$\frac{x^2+3x}{x^2-9}.$$

Find the restrictions from the original denominator, factor, cancel only common factors if they exist, and confirm your final answer still rules out every value the original denominator did. Check the last point carefully: that is exactly where a correct cancellation can still produce a wrong domain.