B10, D1PureStretch

Using Partial Fractions

A rational function with a factorised denominator can be split into partial fractions, each of which is easy to expand binomially. Combining the expansions gives the series for the whole expression.

30 min Sequences and Series

The big picture

Decomposition creates a sum of separate binomial series. Their combined expression is valid only where every series converges. Find each condition from its own scaled bracket before intersecting them. Collect the constant, linear and quadratic terms separately. It is easy to lose a constant when the original rational function has no obvious constant numerator term.

Edexcel A level Maths: (Part 1) 4.3 Using Partial Fractions in Binomial ExpansionCurated 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

Split a rational expression into partial fractions
Rewrite each fraction with a negative power
Expand each part binomially
Combine and state the overall range of validity

The strategy

First decompose into partial fractions. Then write each denominator as a bracket raised to a negative power, e.g. $\tfrac{A}{1-x} = A(1-x)^{-1}$ , and expand each binomially. Add the series term by term.

\frac{f(x)}{(1+ax)(1+bx)} = \frac{A}{1+ax} + \frac{B}{1+bx}

Decompose, then expand each part.

Overall validity

Each expansion has its own range of validity. The combined series is valid only where $\textbf{all}$ of them hold — take the most restrictive (smallest) range.

Tip — Overall validity = the strictest (smallest) of the individual ranges.

Worked example

Expand $\dfrac{3x}{(1+x)(1-2x)}$ .

Example 1Express in partial fractions then find the first two terms.

1Partial fractions:

\tfrac{-1}{1+x} + \tfrac{1}{1-2x}

-(1+x)^{-1} = -(1 - x + \cdots)

;

(1-2x)^{-1} = 1 + 2x + \cdots

3Sum:

(-1+1) + (1+2)x + \cdots = 3x + \cdots

Answer

3x + \cdots

, valid for

|x| < \tfrac12

Intersect the validity ranges

Collect the constant, linear and quadratic terms separately. It is easy to lose a constant when the original rational function has no obvious constant numerator term.

Example 1Expand

\dfrac3{(1-x)(1+2x)}

through

x^2

, stating validity.

1Decomposition gives

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

2Expand:

1+x+x^2+\cdots

and

2(1-2x+4x^2+\cdots)

3Collect to get

3-3x+9x^2+\cdots

4Require

|x|<1

and

|2x|<1

, so the common range is

|x|<1/2

Answer

3-3x+9x^2+\cdots

, for

|x|<1/2

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

Taking the widest range of validity.

The combined series needs every part to converge — use the smallest range.

Expanding before decomposing.

Split into partial fractions first; each part is then a simple binomial.

Key results to remember

\frac{A}{1+ax} = A(1+ax)^{-1}

Negative-power form for expansion.

\text{validity} = \min\text{ of each part}

Strictest range wins.

Key takeaways

Decompose into partial fractions first.
Write each as a negative power and expand binomially.
Combine series; validity is the strictest individual range.

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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