Math Core

A rational expression is a fraction of polynomials, such as $\dfrac{x^2 - 4}{x + 5}$ . You work with these exactly the way you work with numerical fractions: simplify by canceling common factors, multiply straight across, and divide by multiplying by the reciprocal. The one new skill is factoring, and the one new danger is dividing by zero.

Simplifying: cancel factors, not terms

You simplify $\dfrac{12}{18}$ by writing both parts as products and canceling the common factor: $\dfrac{6 \cdot 2}{6 \cdot 3} = \dfrac{2}{3}$ . Rational expressions work the same way, so the first step is always to factor completely.

Simplifying a rational expression

Factor the numerator and the denominator completely.
Note the excluded values: every $x$ that makes the original denominator zero.
Cancel factors that appear in both the numerator and the denominator.

The result is in simplest form when the numerator and denominator have no common factor other than $1$ .

Worked example: Simplifying and excluded values

Simplify $\dfrac{x^2 - 4}{x^2 + 5x + 6}$ and state the excluded values.

Factor both parts:

\frac{x^2 - 4}{x^2 + 5x + 6} = \frac{(x - 2)(x + 2)}{(x + 2)(x + 3)}.

The original denominator is zero at $x = -2$ and $x = -3$ , so those are excluded. Cancel $(x + 2)$ :

\frac{x - 2}{x + 3}, \qquad x \ne -2, -3.

Even though $(x + 2)$ is gone, $x = -2$ is still excluded. The original expression was never defined there. (In graphing terms, that's a hole.)

Common mistake

You may cancel only factors (things multiplied), never terms (things added). In $\dfrac{x + 6}{x + 2}$ you cannot cancel the $x$ 's to get $\dfrac{6}{2} = 3$ . Test with $x = 2$ : the original is $\dfrac{8}{4} = 2$ , not $3$ . If the numerator and denominator aren't written as products, factor first or leave the fraction alone.

Opposite factors

The factors $a - b$ and $b - a$ are opposites: $b - a = -(a - b)$ . So they cancel to $-1$ , not to $1$ .

Worked example: Canceling opposites

Simplify $\dfrac{5 - x}{x^2 - 25}$ .

Factor the denominator and rewrite $5 - x$ as $-(x - 5)$ :

\frac{5 - x}{x^2 - 25} = \frac{-(x - 5)}{(x - 5)(x + 5)} = \frac{-1}{x + 5}, \qquad x \ne 5, -5.

Multiplying

To multiply fractions you multiply numerators and multiply denominators. With rational expressions, it is much easier to factor and cancel before you multiply. Multiplying first creates big polynomials that you'd have to factor again.

\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}

Worked example: Multiplying rational expressions

Multiply $\dfrac{x + 3}{x - 1} \cdot \dfrac{x^2 - 1}{x^2 + 6x + 9}$ .

Factor everything:

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

Now cancel one $(x + 3)$ and the $(x - 1)$ . Any factor in any numerator may cancel with a matching factor in any denominator:

= \frac{x + 1}{x + 3}, \qquad x \ne 1, -3.

Dividing

Dividing by a fraction means multiplying by its reciprocal:

\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}.

Flip only the second fraction, then multiply as before.

Worked example: Dividing rational expressions

Divide $\dfrac{x^2 - 25}{2x} \div \dfrac{x + 5}{4x^2}$ .

Flip the divisor and factor:

\frac{(x - 5)(x + 5)}{2x} \cdot \frac{4x^2}{x + 5}.

Cancel $(x + 5)$ , and simplify $\dfrac{4x^2}{2x} = 2x$ :

= 2x(x - 5), \qquad x \ne 0, -5.

Why is $x = -5$ excluded? It made the numerator of the divisor zero, and dividing by $\dfrac{0}{4x^2} = 0$ is undefined.

In a division problem, the excluded values come from three places: the denominator of the first fraction, the denominator of the second fraction, and the numerator of the second fraction (because it becomes a denominator after flipping).

Tip

Check any simplification by plugging in a number that is not excluded. For the division above, try $x = 1$ . Original: $\dfrac{-24}{2} \div \dfrac{6}{4} = -12 \cdot \dfrac{4}{6} = -8$ . Simplified: $2(1)(1 - 5) = -8$ . They agree.

Practice

Practice 1

Simplify $\dfrac{x^2 - 9}{x^2 - x - 6}$ .

Practice 2

What are the excluded values of $\dfrac{x^2 - 9}{x^2 - x - 6}$ ?

Separate answers with commas, e.g. 2, -5

Practice 3

Multiply $\dfrac{x + 4}{x - 2} \cdot \dfrac{x^2 - 4}{x^2 + 8x + 16}$ .

Practice 4

Simplify $\dfrac{7 - x}{x^2 - 49}$ .

Practice 5

Divide $\dfrac{x^2 - 16}{3x} \div \dfrac{x - 4}{6x^3}$ .

Practice 6

The quotient $\dfrac{x^2 + x - 12}{x^2 - 9} \div \dfrac{x + 4}{x + 3}$ simplifies to a constant (for all allowed $x$ ). What is that constant?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Find every value of $x$ excluded from $\dfrac{x + 2}{x - 1} \div \dfrac{x - 3}{x + 4}$ .

Separate answers with commas, e.g. 2, -5

Practice 8

Simplify $\dfrac{2x^2 + 5x - 3}{x^2 - 9} \cdot \dfrac{x - 3}{4x - 2}$ .