1.02yPureStretch

Repeated Factors

When a denominator has a repeated linear factor like $(x+a)^2$ , 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 Polynomials

The big picture

For a squared linear factor and one distinct linear factor, include all three permitted terms. Values that zero factors reveal some constants quickly; a remaining coefficient equation reveals the last one. Verify by recombining the numerator, especially if the same expression will be used for integration or a binomial expansion. Preserve the excluded denominator values.

Edexcel A level Maths: 1.4 Repeated Factors (Partial Fractions)Curated by Kepler, not produced by us. Watch it, then work through 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

The extra term

For OCR H240, a squared linear factor contributes two terms: one over the factor and one over its square. For a denominator $(x+a)^2(x+b)$ , include $A/(x+a)+B/(x+a)^2+C/(x+b)$ . Omitting the lower-power term removes a necessary degree of freedom.

\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.

Why the extra term is needed

Without the $\dfrac{B}{(x+a)^2}$ 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.

A complete decomposition with a squared factor

Verify by recombining the numerator, especially if the same expression will be used for integration or a binomial expansion. Preserve the excluded denominator values.

Example 1Decompose

\dfrac{3x+1}{(x-1)(x+1)^2}

1Write

A/(x-1)+B/(x+1)+C/(x+1)^2

2Clearing the denominator gives

3x+1=A(x+1)^2+B(x-1)(x+1)+C(x-1)

3At x = 1: A = 1. At x = -1: C = 1. Comparing quadratic coefficients gives A+B = 0, so B = -1.

4Check:

(x+1)^2-(x-1)(x+1)+(x-1)=3x+1

Answer

\dfrac1{x-1}-\dfrac1{x+1}+\dfrac1{(x+1)^2}

, with

x\neq\pm1

Example 2Decompose

\dfrac1{x(x+1)^2}

and check it by recombining.

1Write

A/x+B/(x+1)+C/(x+1)^2

, then

1=A(x+1)^2+Bx(x+1)+Cx

2At x = 0, A = 1; at x = -1, C = -1. The quadratic coefficient gives B = -1.

3Recombining gives

(x+1)^2-x(x+1)-x=1

Answer

1/x-1/(x+1)-1/(x+1)^2

, for

x\neq0,-1

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

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 results to remember

\dfrac{A}{x+a} + \dfrac{B}{(x+a)^2}

Repeated linear factor (squared).

\text{compare coefficients for the last constant}

Finishing off.

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.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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