Multiplying Rational Functions

Multiplying rational functions works like multiplying any fractions — multiply numerators together and denominators together. But with polynomial expressions, there's almost always a better path: factor first, cancel, then multiply. This avoids expanding large polynomials that would just get simplified away anyway.

The Method

  1. Factor every numerator and denominator completely.
  2. Cancel any factor that appears in both a numerator and a denominator.
  3. Multiply what's left.

Keep track of domain restrictions — any value that made a denominator zero in the original problem remains excluded, even after canceling.

Examples

Example 1

$$\frac{x+1}{x+3} \cdot \frac{2x+3}{x-1}$$

Factor each piece: none of these factor further. No common factors to cancel, so just multiply:

$$\frac{(x+1)(2x+3)}{(x+3)(x-1)} = \frac{2x^2 + 5x + 3}{x^2 + 2x - 3}$$

You can leave the answer in factored form or expand — factored is often preferred since it makes any remaining simplification easier to spot.

Example 2

$$\frac{x^2 - 4}{x - 3} \cdot \frac{x^2 - 7x + 12}{x^2 - 2x}$$

Step 1 — Factor everything:

  • \(x^2 - 4 = (x-2)(x+2)\)
  • \(x - 3\) is already factored
  • \(x^2 - 7x + 12 = (x-3)(x-4)\)
  • \(x^2 - 2x = x(x-2)\)

Rewrite with all factors visible:

$$\frac{(x-2)(x+2)}{(x-3)} \cdot \frac{(x-3)(x-4)}{x(x-2)}$$

Step 2 — Cancel common factors:

  • \((x-2)\) appears in a numerator and denominator — cancel.
  • \((x-3)\) appears in a numerator and denominator — cancel.

$$\frac{\cancel{(x-2)}(x+2)}{\cancel{(x-3)}} \cdot \frac{\cancel{(x-3)}(x-4)}{x\cancel{(x-2)}} = \frac{(x+2)(x-4)}{x}$$

Step 3 — Multiply what's left:

$$\frac{(x+2)(x-4)}{x} = \frac{x^2 - 2x - 8}{x}$$

Domain restrictions from the original: \(x \neq 0, 2, 3\).

Example 3

$$\frac{3x^2 + 6x}{x^2 - 1} \cdot \frac{x^2 - x - 2}{x^2 + 4x + 4}$$

Factor:

  • \(3x^2 + 6x = 3x(x+2)\)
  • \(x^2 - 1 = (x-1)(x+1)\)
  • \(x^2 - x - 2 = (x-2)(x+1)\)
  • \(x^2 + 4x + 4 = (x+2)^2\)

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

Cancel: \((x+1)\) cancels, one \((x+2)\) cancels:

$$\frac{3x\cancel{(x+2)}}{(x-1)\cancel{(x+1)}} \cdot \frac{(x-2)\cancel{(x+1)}}{\cancel{(x+2)}(x+2)} = \frac{3x(x-2)}{(x-1)(x+2)}$$

Domain: \(x \neq 0, -1, 1, -2\).

Practice Problems

1. Multiply: \(\dfrac{x^2-1}{x+3} \cdot \dfrac{x+3}{x-1}\) Show answerFactor: \(x^2-1 = (x-1)(x+1)\). The \((x+3)\) and \((x-1)\) each cancel: \(\frac{(x-1)(x+1)\cancel{(x+3)}}{\cancel{(x+3)}\cancel{(x-1)}} = x+1\), with \(x \neq 1, -3\).

2. Multiply: \(\dfrac{x^2+3x+2}{x^2-4} \cdot \dfrac{x-2}{x+1}\) Show answerFactor: \(x^2+3x+2 = (x+1)(x+2)\), \(x^2-4 = (x-2)(x+2)\). Cancel \((x+2)\), \((x-2)\), and \((x+1)\): result is \(1\), with \(x \neq 2, -2, -1\).

3. Multiply: \(\dfrac{2x^2-8}{x^2+x-6} \cdot \dfrac{x-2}{x^2-4x+4}\) Show answerFactor: \(2x^2-8 = 2(x-2)(x+2)\), \(x^2+x-6 = (x+3)(x-2)\), \(x^2-4x+4 = (x-2)^2\). After canceling \((x-2)\) twice: \(\frac{2(x+2)}{(x+3)(x-2)}\), with \(x \neq 2, -3, -2\).

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