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1.4PureStretch

Repeated Factors

When a denominator has a repeated linear factor like (x+a)2(x+a)^2, partial fractions need an extra term — one for each power up to the highest. The constants are then found by a mix of substitution and comparing coefficients.

30 min Algebraic Methods
Edexcel A level Maths: 1.4 Repeated Factors (Partial Fractions)Curated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Set up partial fractions with a repeated factor
  • Include a term for each power of the repeated factor
  • Find the constants by substitution and comparing coefficients
  • Combine repeated and distinct factors
1

The extra term

A repeated factor (x+a)2(x + a)^2 contributes two\textbf{two} partial fractions: one over (x+a)(x + a) and one over (x+a)2(x + a)^2. In general (x+a)n(x+a)^n needs nn terms, with powers 11 up to nn.

…(x+a)2(x+b)=Ax+a+B(x+a)2+Cx+b\dfrac{\ldots}{(x + a)^2 (x + b)} = \dfrac{A}{x + a} + \dfrac{B}{(x + a)^2} + \dfrac{C}{x + b}
One term per power of the repeated factor.
2

Finding the constants

Clear the denominator. Substituting the roots gives the constants over the highest power and over the distinct factor directly; the remaining constant comes from comparing coefficients\textbf{comparing coefficients} (or substituting one more value).

Example 1Outline finding A, B, C for …(x+1)2(x−2)\dfrac{\ldots}{(x+1)^2(x-2)}.
1Substitute x=−1x = -1 to find BB (over (x+1)2(x+1)^2).
2Substitute x=2x = 2 to find CC (over x−2x - 2).
3Compare coefficients (e.g. of x2x^2) or substitute another xx to find AA.
Answer
B and C by roots, A by comparing coefficients

Tip — Substitution gives the “easy” constants; the one over the LOWER power of the repeat usually needs comparing coefficients.

3

Why the extra term is needed

Without the B(x+a)2\dfrac{B}{(x+a)^2} term, the partial fractions could not recombine to the original — there would not be enough freedom. Each power of the repeated factor is genuinely needed.

Formula recap

Ax+a+B(x+a)2\dfrac{A}{x+a} + \dfrac{B}{(x+a)^2}
Repeated linear factor (squared).
(x+a)n⇒n terms(x+a)^n \Rightarrow n \text{ terms}
Powers 1 … n.
compare coefficients for the last constant\text{compare coefficients for the last constant}
Finishing off.

Common mistakes to avoid

Using only one term for a repeated factor.
(x + a)² needs BOTH A/(x+a) and B/(x+a)².
Expecting substitution alone to find every constant.
The constant over the lower power usually needs comparing coefficients.

Key takeaways

  • A repeated factor (x+a)² adds a term over (x+a)².
  • (x+a)ⁿ needs n terms, with powers 1 up to n.
  • Find constants by substituting roots, then comparing coefficients for the rest.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. A factor (x+a)2(x+a)^{2} in the denominator needs how many partial-fraction terms?

    • Two
    • One
    • Three
    • Zero
    Show answer and explanation

    Answer: Two

    Over (x+a) and (x+a)².

  2. The terms for a repeated factor (x+a)2(x+a)^{2} are over:

    • (x+a)(x+a) and (x+a)2(x+a)^{2}
    • (x+a)2(x+a)^{2} only
    • (x+a)(x+a) only
    • (x+a)3(x+a)^{3}
    Show answer and explanation

    Answer: (x+a)(x+a) and (x+a)2(x+a)^{2}

    Powers 1 and 2.

  3. (x+a)3(x+a)^{3} would require how many terms?

    • Three
    • Two
    • One
    • Four
    Show answer and explanation

    Answer: Three

    Powers 1, 2, 3.

  4. Substituting a root typically gives the constant over the:

    • Highest power / distinct factor
    • Lowest power
    • Numerator
    • Whole fraction
    Show answer and explanation

    Answer: Highest power / distinct factor

    Those isolate cleanly.

  5. The remaining constant is usually found by:

    • Comparing coefficients
    • Guessing
    • Differentiating
    • Cover-up
    Show answer and explanation

    Answer: Comparing coefficients

    Or substituting another value.

  6. Omitting the (x+a)² term means the partial fractions:

    • Cannot recombine to the original
    • Are simpler but correct
    • Have too many constants
    • Are improper
    Show answer and explanation

    Answer: Cannot recombine to the original

    Not enough freedom without it.

Next step

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