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MathidiaMathematics Academy
Partial fractions
Lesson 6 of 75
Introduction

Distinct linear factors

Objective. Resolve proper fractions into simple terms.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Distinct linear factors in partial fractions let a proper rational expression with denominator (x−a)(x−b) be rewritten as A/(x−a)+B/(x−b), with A and B constants found by substitution or comparing coefficients. Example: 7x−3/((x−1)(x+2)) splits as A/(x−1)+B/(x+2); multiplying through and substituting x=1 gives A=4/3, and x=−2 gives B=17/3. This decomposition is essential before integrating or expanding such fractions as series.

Worked example

Express 5x+1/((x−2)(x+3)) in partial fractions with distinct linear factors.

5 … 1
Illustrated card showing 5 … 1
  1. 1Write 5x+1/((x−2)(x+3))=A/(x−2)+B/(x+3)
  2. 2Multiply both sides by (x−2)(x+3) to get 5x+1=A(x+3)+B(x−2)
  3. 3Substitute x=2: 11=5A, so A=11/5; substitute x=−3: −14=−5B, so B=14/5
  4. 4Write the final decomposition using these values

Answer: 11/(5(x−2))+14/(5(x+3))

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Common mistake. Students forget the expression must be a proper fraction (numerator degree less than denominator) before decomposing; applying the method directly to an improper fraction gives an incorrect split.

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