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Partial Fractions Decomposition

A free College Algebra lesson from the “Rational Expressions and Equations” unit, with a worked example and practice problems including step-by-step solutions.

Partial fractions decomposition rewrites a complex rational expression as a sum of simpler ones. For a proper rational P(x) / [(x - a)(x - b)] with distinct linear factors, write P(x) / [(x - a)(x - b)] = A / (x - a) + B / (x - b). Multiply both sides by the common denominator and solve for A and B (often by plugging in x = a and x = b).

What you'll learn

Why it matters: Calculus integration of rational functions, Laplace transforms in engineering, and any closed-form solution of differential equations all use partial fractions decomposition.

Worked example

Problem. Decompose 1 / [(x - 1)(x - 2)] into A / (x - 1) + B / (x - 2). Find A.

  1. Multiply both sides by (x - 1)(x - 2): 1 = A(x - 2) + B(x - 1).
  2. Set x = 1: 1 = A(1 - 2) = -A, so A = -1.
  3. Set x = 2: 1 = B(2 - 1) = B, so B = 1.

Answer: -1

Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. Decompose 1 / [(x - 1)(x - 2)] = A / (x - 1) + B / (x - 2). Find A.

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  1. Multiply both sides by denominator. Set x = 1: 1 = -A so A = -1.

Answer: -1

2. Same. Find B.

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  1. Set x = 2: 1 = B.

Answer: 1

3. Decompose 5 / [(x - 1)(x - 2)] = A / (x - 1) + B / (x - 2). Find A.

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  1. Same method: set x = 1, get 5 = -A -> A = -5.

Answer: -5

4. Same. Find B.

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  1. Set x = 2: 5 = B.

Answer: 5

5. Decompose 1/(x(x - 1)) = A/x + B/(x - 1). Find A.

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  1. Set x = 0: 1 = A(0 - 1) = -A, so A = -1.

Answer: -1

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