1.4PureStretch

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 Video by Zeeshan Zamurred Algebraic Methods

Edexcel A level Maths: 1.4 Repeated Factors (Partial Fractions)Watch the full walkthrough 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

The extra term

A repeated factor $(x + a)^{2}$ contributes $two$ partial fractions: one over $(x + a)$ and one over $(x + a)^{2}$ . In general $(x + a)^{n}$ needs $n$ terms, with powers $1$ up to $n$ .

\frac{\dots}{( x + a ) ^{2} ( x + b )} = \frac{A}{x + a} + \frac{B}{( x + a ) ^{2}} + \frac{C}{x + b}

One term per power of the repeated factor.

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 $comparing coefficients$ (or substituting one more value).

1Substitute

x = - 1

to find

B

(over

(x + 1)^{2}

2Substitute

x = 2

to find

C

(over

x - 2

3Compare coefficients (e.g. of

x^{2}

) or substitute another

x

to find

A

AnswerB 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.

Why the extra term is needed

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

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

Repeated linear factor (squared).

(x + a)^{n} \Rightarrow n terms

Powers 1 … n.

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

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