Rational Expressions — AP® Precalculus Notes

Rational expressions are fractions with polynomials on top and bottom. Topic 1.7 builds functions out of them, and their behavior depends on details of the algebra: which factors cancel and which inputs are excluded. This page is that algebra.

What a rational expression is

A rational expression is a quotient $\frac{p(x)}{q(x)}$ of two polynomials, such as $\frac{x - 3}{x - 2}$ or $\frac{2x^2 + 3}{x}$ . It is undefined wherever the denominator $q(x)$ is $0$ . Find those inputs by factoring the denominator: $\frac{x + 2}{x^2 - 4x}$ $= \frac{x + 2}{x(x - 4)}$ is undefined at $x = 0$ and $x = 4$ .

Simplifying

Factor the numerator and denominator completely, then cancel factors they share.

Worked example 1. Simplify $\frac{x^2 - 9}{x^2 + x - 6}$ .

Factor: $\frac{(x - 3)(x + 3)}{(x + 3)(x - 2)}$ .

Cancel $x + 3$ : $\frac{x - 3}{x - 2}$ , with $x \ne -3$ and $x \ne 2$ .

The excluded input $x = -3$ does not come back when the factor cancels. The original expression has no value at $x = -3$ , so the simplified one is equal to it only for $x \ne -3$ . On a graph, that shows up as a single missing point.

The graph of y = (x² − 4)/(x − 2) is the line y = x + 2 with one point missing: an open circle at (2, 4), because x = 2 is not in the domain of the original expression. — **Figure 1.** $\frac{x^2 - 4}{x - 2}$ agrees with $x + 2$ everywhere except $x = 2$ . LevlPrep original diagram.

Cancel factors, never terms. In $\frac{x + 6}{x + 3}$ the $x$ 's are terms of sums; nothing cancels. In $\frac{x^2 + x}{x}$ $= \frac{x(x + 1)}{x}$ , $x$ is a factor of the whole top, so it cancels to $x + 1$ for $x \ne 0$ .

Adding and subtracting

Use a common denominator, the product of the different factors.

Worked example 2. Write $\frac{1}{x - 1} + \frac{2}{x + 1}$ as one fraction.

Common denominator $(x - 1)(x + 1)$ . Multiply each numerator by the factor its denominator lacks:

$\frac{1(x + 1) + 2(x - 1)}{(x - 1)(x + 1)}$ $= \frac{3x - 1}{x^2 - 1}$ .

Multiplying and dividing

Multiply numerators and denominators straight across, cancelling common factors. To divide, multiply by the reciprocal of the second fraction:

$\frac{x^2 - 1}{x + 3} \div \frac{x - 1}{x^2 - 9} = \frac{(x - 1)(x + 1)}{x + 3} \cdot \frac{(x - 3)(x + 3)}{x - 1} = (x + 1)(x - 3).$

A fraction inside a fraction simplifies the same way: multiply its top and bottom by the small denominators. $\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$ $= \frac{x + 1}{x - 1}$ .

Splitting a fraction

The reverse of adding: when the numerator is a sum, divide each term by the denominator. $\frac{2x^2 + 3}{x} = 2x + \frac{3}{x}$ . When the denominator has two terms, first write the numerator in terms of it: $x^2 + 2x + 3$ $= (x + 1)^2 + 2$ , so

$\frac{x^2 + 2x + 3}{x + 1} = x + 1 + \frac{2}{x + 1}.$

Topic 1.7 uses this form: for large $x$ , the small fraction $\frac{2}{x + 1}$ is close to $0$ , so the expression behaves like $x + 1$ .