Repeated linear factors
Objective. Use the complete decomposition for repeated roots.
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- 2. Worked example
- 3. Practice
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Learn: the key idea
Repeated linear factors appear when a denominator contains (x−a)² or higher, requiring an extra term for each power, such as A/(x−a)+B/(x−a)² for a squared factor. Example: 3x+1/(x−1)²(x+2) decomposes as A/(x−1)+B/(x−1)²+C/(x+2), with each unknown found by substitution and comparing coefficients. Missing the extra power term leaves the decomposition incomplete and unable to reproduce the original fraction when recombined.
Worked example
Decompose 4x+5/(x+1)²(x−2) using partial fractions with a repeated factor.
- 1Write 4x+5/(x+1)²(x−2)=A/(x+1)+B/(x+1)²+C/(x−2)
- 2Multiply through: 4x+5=A(x+1)(x−2)+B(x−2)+C(x+1)²
- 3Substitute x=−1: 1=−3B, so B=−1/3; substitute x=2: 13=9C, so C=13/9
- 4Compare constant terms to find A: 5=−2A−2B+C, giving A=−13/9
Answer: −13/(9(x+1))−1/(3(x+1)²)+13/(9(x−2))
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Start practisingCommon mistake. Students write only one term A/(x−1)² for a repeated factor, omitting the necessary B/(x−1) term, which makes the decomposition impossible to balance correctly.
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