B12PureStretch

Partial Fractions

Partial fractions run the addition of fractions in reverse: they break a single complicated fraction back into a sum of simple ones. It is a tidying tool that makes integration and binomial expansion possible later on.

30 min Video by TLMaths Algebra and Functions

The big picture

You already know how to add $\frac{1}{x - 1} + \frac{2}{x + 3}$ by finding a common denominator, producing one messy fraction. Partial fractions ask the reverse question: given the messy fraction, can you recover the simple pieces it came from? On its own that looks like busywork — but those simple pieces are exactly what you need to integrate a rational function or to expand it as a binomial series. So this lesson is an investment: it is quietly a prerequisite for two big Year-2 techniques, and the algebra (factorising, the factor theorem, solving for unknowns) ties the whole chapter together.

A-Level Maths B10-03: Introducing Partial Fractions via SubstitutionWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Recognise when an expression can be split into partial fractions
Split a fraction with distinct linear factors
Use substitution (cover-up) or comparing coefficients to find the constants
Handle a repeated linear factor

The idea: adding fractions, backwards

Adding $\frac{A}{x - p} + \frac{B}{x - q}$ gives a single fraction whose denominator is $(x - p) (x - q)$ . $Partial fractions$ reverse this: start from a fraction with a factorised denominator and find the constants $A$ and $B$ that rebuild it.

For this to work the fraction must be $proper$ — the numerator’s degree lower than the denominator’s. If it is not, divide first (as in the polynomials lesson) to split off a polynomial part, then apply partial fractions to the remaining proper fraction.

\frac{numerator}{( x - p ) ( x - q )} = \frac{A}{x - p} + \frac{B}{x - q}

One fraction per distinct linear factor.

Finding the constants by substitution

The fastest method multiplies through by the whole denominator to clear fractions, then $substitutes clever values$ of $x$ — each chosen to make one bracket zero, so only one unknown survives. This is often called the “cover-up” method.

1Write

\frac{5 x - 1}{( x - 1 ) ( x + 2 )} = \frac{A}{x - 1} + \frac{B}{x + 2}

2Multiply through by

(x - 1) (x + 2)

5 x - 1 = A (x + 2) + B (x - 1)

3Let

x = 1

(kills

B

5 (1) - 1 = A (3) \Rightarrow 4 = 3 A \Rightarrow A = \frac{4}{3}

4Let

x = - 2

(kills

A

5 (- 2) - 1 = B (- 3) \Rightarrow - 11 = - 3 B \Rightarrow B = \frac{11}{3}

Answer

\frac{5 x - 1}{( x - 1 ) ( x + 2 )} = \frac{4}{3 ( x - 1 )} + \frac{11}{3 ( x + 2 )}

Why this works

Substituting the value that makes a bracket zero is the whole trick: it switches off every term but one, so each constant falls out on its own instead of solving simultaneous equations.

Comparing coefficients

An alternative — useful as a check or when substitution is awkward — is to expand the right-hand side and $match coefficients$ of each power of $x$ with the left. This gives simultaneous equations for the constants. Either method is valid; substitution is usually quicker for linear factors.

1From

5 x - 1 = A (x + 2) + B (x - 1)

, expand:

5 x - 1 = (A + B) x + (2 A - B)

2Match

x

-terms:

A + B = 5

. Match constants:

2 A - B = - 1

3Add the equations:

3 A = 4 \Rightarrow A = \frac{4}{3}

, then

B = 5 - \frac{4}{3} = \frac{11}{3}

AnswerSame result:

A = \frac{4}{3}

B = \frac{11}{3}

Repeated factors

When a factor is repeated, such as $(x - 2)^{2}$ , it needs $two$ fractions — one over $(x - 2)$ and one over $(x - 2)^{2}$ . In general a factor to the power $n$ contributes fractions with denominators up to that power. Miss one and the decomposition cannot be made to fit.

\frac{\dots}{( x - p ) ^{2}} = \frac{A}{x - p} + \frac{B}{( x - p ) ^{2}}

A squared factor needs both a first- and second-power term.

Tip — Count fractions before you start: one per distinct linear factor, plus an extra for each repeat. Getting the $form$ right is half the battle.

Think like an examiner

State the correct form first — one fraction per distinct factor, and an extra term for each repeated factor. Marks are awarded for the setup.
Substituting the root of each bracket ( $x = p$ ) is the quickest route to the constants; show the substitution clearly.
Check the fraction is proper; if the top degree is $\geq$ the bottom, divide out a polynomial part first.
A final answer can be verified by adding your partial fractions back together — a quick self-check worth doing.

Common misconceptions

Using one fraction for a repeated factor

(x - 2)^{2}

A squared factor needs two:

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

Applying partial fractions to an improper fraction directly.

Divide first to extract the polynomial part, then decompose the proper remainder.

Forgetting to multiply the whole equation through by the denominator before substituting.

Clear the fractions first, giving a polynomial identity, then substitute values.

Substituting a value that does not simplify anything.

Choose values that make a bracket zero, so only one unknown remains each time.

Partial fractions forms

\frac{p x + q}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}

Distinct linear factors.

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

Repeated factor.

degree(top) < degree(bottom)

Must be a proper fraction.

Stretch yourself

Express $\frac{11 - 3 x}{( x - 3 ) ( x + 1 ) ^{2}}$ in partial fractions.

Hint — The repeated factor needs two terms: use $\frac{A}{x - 3} + \frac{B}{x + 1} + \frac{C}{( x + 1 ) ^{2}}$ . Substitute $x = 3$ and $x = - 1$ to get $A$ and $C$ , then compare coefficients (or substitute another value) for $B$ .

Questions students ask

What are partial fractions used for?

They split a complicated rational expression into simple pieces that are much easier to integrate or to expand as a binomial series. On their own they are a tidying step; their real value shows up in Year-2 calculus and series work.

How many fractions do I need?

One for each distinct linear factor in the denominator, plus an extra term for each repeated factor. A factor

(x - a)^{2}

contributes both

\frac{A}{x - a}

and

\frac{B}{( x - a ) ^{2}}

Which method is best for finding the constants?

Substituting the value that makes each bracket zero (the cover-up method) is usually fastest for linear factors, because it isolates one constant at a time. Comparing coefficients is a reliable alternative and a good check.

What if the fraction is improper?

If the numerator’s degree is at least the denominator’s, divide first to pull out a polynomial part, then apply partial fractions to the proper remainder. You cannot decompose an improper fraction directly.

Key takeaways

Partial fractions reverse the adding of fractions, recovering the simple pieces.
Use one fraction per distinct linear factor; an extra term for each repeat.
Clear the denominator, then substitute the root of each bracket to isolate a constant.
Comparing coefficients is an alternative and a useful check.
The fraction must be proper — divide first if it is not.

How this fits the course

Build on

Integration (later)soon
Binomial Expansion (later)soon

Leads to

Coordinate Geometry

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