Lesson 4 / 5 · 12 min read

Rational Expressions

Simplify, combine, and solve — and don't forget the domain.

What is a rational expression?

A rational expression is a fraction whose top and bottom are polynomials.

$\dfrac{x^2 - 4}{x^2 + 3x + 2}$

Treat them like numerical fractions, but watch the domain: any $x$ that makes the denominator zero is excluded.

Factor top and bottom; cancel common factors.

Example. $\dfrac{x^2 - 4}{x^2 + 3x + 2}$ :

$\dfrac{(x-2)(x+2)}{(x+1)(x+2)} = \dfrac{x - 2}{x + 1}, \quad x \neq -2,\ -1$

Even though $(x+2)$ cancels, we still exclude $x = -2$ — the original expression isn't defined there.

Multiply straight across; factor first to cancel.

$\dfrac{x + 1}{x - 3} \cdot \dfrac{x - 3}{x^2 - 1} = \dfrac{x + 1}{x - 3} \cdot \dfrac{x - 3}{(x-1)(x+1)} = \dfrac{1}{x - 1}$

Divide by multiplying by the reciprocal.

$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c}$

You need a common denominator. The least common denominator (LCD) is the simplest one that works.

Example. $\dfrac{2}{x} + \dfrac{3}{x + 1}$ . LCD: $x(x + 1)$ .

$\dfrac{2(x+1)}{x(x+1)} + \dfrac{3x}{x(x+1)} = \dfrac{5x + 2}{x(x+1)}$

Multiply both sides by the LCD to clear fractions. Then check that no candidate solution is excluded by the domain.

Example. Solve $\dfrac{1}{x} + \dfrac{2}{3} = 1$ .

Multiply both sides by $3x$ :

$3 + 2x = 3x \implies x = 3$

Check: $x = 3$ doesn't make any denominator zero, so it's valid.

Extraneous solution warning. Multiplying by the LCD can introduce a value that doesn't satisfy the original equation. Always check.