Partial Fractions in Depth

Builds on: aqa-alevel-maths-pure-1 / algebra-and-functions — this lesson is the A2-level deep dive.

This lesson covers partial fractions in depth as required by the AQA A-Level Mathematics specification (7357). Partial fractions allow you to decompose a single algebraic fraction into a sum of simpler fractions. This technique is essential for integration, for summing series, and for working with binomial expansions of rational functions.

Spec Mapping — AQA 7357 Section B Algebra and Functions. This lesson covers the partial-fraction decomposition content of Section B, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

Why Partial Fractions?

Many algebraic fractions are difficult to integrate or manipulate in their combined form. By splitting them into simpler fractions, we can:

Integrate expressions that would otherwise be intractable.
Sum series using the method of differences (telescoping).
Expand rational functions using the binomial theorem.

For example, it is not obvious how to integrate $\dfrac{1}{(x - 1)(x + 2)}$ , but if we write it as:

\frac{1}{(x - 1)(x + 2)} = \frac{1/3}{x - 1} - \frac{1/3}{x + 2}

then each term can be integrated using the standard result $\int \dfrac{1}{x - a}\,dx = \ln\lvert x - a\rvert + C$ .

Case 1: Distinct Linear Factors

If the denominator has distinct linear factors, the partial fraction decomposition has the form:

\frac{f(x)}{(x - a)(x - b)(x - c)} = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}

Method: Cover-Up Rule (Heaviside)

To find $A$ , "cover up" $(x - a)$ in the original fraction and substitute $x = a$ .

Example: Express $\dfrac{3x + 5}{(x + 1)(x - 2)}$ in partial fractions.

\frac{3x + 5}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}

Cover up $(x + 1)$ and set $x = -1$ : $A = \dfrac{3(-1) + 5}{-1 - 2} = \dfrac{2}{-3} = -\tfrac{2}{3}$ .

Cover up $(x - 2)$ and set $x = 2$ : $B = \dfrac{3(2) + 5}{2 + 1} = \tfrac{11}{3}$ .

So $\dfrac{3x + 5}{(x + 1)(x - 2)} = \dfrac{-2/3}{x + 1} + \dfrac{11/3}{x - 2}$ .

Alternative method: Equating Coefficients

Multiply both sides by $(x + 1)(x - 2)$ :

3x + 5 = A(x - 2) + B(x + 1)

Expand: $3x + 5 = (A + B)x + (-2A + B)$ .

Equating coefficients:

$x$ : $A + B = 3$
constant: $-2A + B = 5$

Solving: from the first equation $B = 3 - A$ ; substituting: $-2A + 3 - A = 5$ , so $-3A = 2$ , $A = -\tfrac{2}{3}$ , $B = \tfrac{11}{3}$ .

Example: Express $\dfrac{5x^2 + 3x + 1}{x(x + 1)(x - 1)}$ in partial fractions.

= \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}

Cover-up:

$x = 0$ : $A = \dfrac{0 + 0 + 1}{(0 + 1)(0 - 1)} = \dfrac{1}{-1} = -1$ .
$x = -1$ : $B = \dfrac{5 - 3 + 1}{(-1)(-1 - 1)} = \dfrac{3}{2} = \tfrac{3}{2}$ .
$x = 1$ : $C = \dfrac{5 + 3 + 1}{(1)(1 + 1)} = \tfrac{9}{2}$ .

So $\dfrac{5x^2 + 3x + 1}{x(x + 1)(x - 1)} = -\dfrac{1}{x} + \dfrac{3/2}{x + 1} + \dfrac{9/2}{x - 1}$ .

Case 2: Repeated Linear Factors

If the denominator contains a repeated factor (x − a)², the decomposition includes:

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

Example: Express $\dfrac{4x + 1}{(x - 1)^2(x + 2)}$ in partial fractions.

\frac{4x + 1}{(x - 1)^{2}(x + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{x + 2}

Multiply through by $(x - 1)^2(x + 2)$ :

4x + 1 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^{2}

Substituting strategic values:

$x = 1$ : $5 = B(3)$ , so $B = \tfrac{5}{3}$ . $x = -2$ : $-7 = C(-3)^2 = 9C$ , so $C = -\tfrac{7}{9}$ .

Equating $x^2$ coefficients: $0 = A + C$ , so $A = \tfrac{7}{9}$ .

Result: $\dfrac{4x + 1}{(x - 1)^2(x + 2)} = \dfrac{7/9}{x - 1} + \dfrac{5/3}{(x - 1)^2} + \dfrac{-7/9}{x + 2}$ .

Case 3: Improper Fractions

A fraction is improper if the degree of the numerator is greater than or equal to the degree of the denominator. Before decomposing, you must perform polynomial long division to obtain a polynomial plus a proper fraction.

\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}

where the degree of $r(x)$ is less than the degree of $g(x)$ .

Example: Express $\dfrac{x^3 + 2x^2 - x + 3}{(x + 1)(x - 2)}$ in partial fractions.

The denominator expands to $x^2 - x - 2$ (degree 2), and the numerator has degree 3, so the fraction is improper.

Divide $x^3 + 2x^2 - x + 3$ by $x^2 - x - 2$ :

x^{3} + 2x^{2} - x + 3 = (x + 3)(x^{2} - x - 2) + (4x + 9)

Check: $(x + 3)(x^2 - x - 2) = x^3 - x^2 - 2x + 3x^2 - 3x - 6 = x^3 + 2x^2 - 5x - 6$ . Then $x^3 + 2x^2 - x + 3 - (x^3 + 2x^2 - 5x - 6) = 4x + 9$ . ✓

So $\dfrac{x^3 + 2x^2 - x + 3}{(x + 1)(x - 2)} = x + 3 + \dfrac{4x + 9}{(x + 1)(x - 2)}$ .

Now decompose the proper fraction:

\frac{4x + 9}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}

$x = -1$ : $A = \dfrac{-4 + 9}{-3} = -\tfrac{5}{3}$ . $x = 2$ : $B = \dfrac{8 + 9}{3} = \tfrac{17}{3}$ .

Final answer: $x + 3 - \dfrac{5/3}{x + 1} + \dfrac{17/3}{x - 2}$ .

Applications

Application to Integration

\begin{aligned} \int \frac{3x + 5}{(x + 1)(x - 2)}\,dx &= \int \frac{-2/3}{x + 1}\,dx + \int \frac{11/3}{x - 2}\,dx\\ &= -\tfrac{2}{3}\ln|x + 1| + \tfrac{11}{3}\ln|x - 2| + C \end{aligned}

Application to Series (Method of Differences)

Example: Show that $\dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r + 1}$ . Hence find $\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 1)}$ .

The partial fraction decomposition gives $\dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r + 1}$ (cover-up rule).

\begin{aligned} \sum_{r=1}^{n} \left[\frac{1}{r} - \frac{1}{r + 1}\right] &= \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{4}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n + 1}\right)\\ &= 1 - \frac{1}{n + 1}\\ &= \frac{n}{n + 1} \end{aligned}

This is the method of differences (telescoping series).

Synoptic Links

This topic connects to:

Integration techniques (Pure 2) (aqa-alevel-maths-pure-2 / integration-techniques) — partial fractions are the gateway to rational integration giving ln-type integrals.
Integration by substitution (aqa-alevel-maths-calculus-applications / integration-by-substitution) — many substitution problems reduce to a partial-fraction step.
Algebraic fractions and partial fractions (Pure 2) (aqa-alevel-maths-pure-2 / algebraic-fractions-and-partial-fractions) — the foundational Pure 2 treatment this deep dive extends.

Summary

Distinct linear factors: decompose as $\dfrac{A}{x - a} + \dfrac{B}{x - b} + \cdots$
Repeated linear factors: include terms $\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2} + \cdots$
Improper fractions: divide first, then decompose the proper remainder.
Use the cover-up rule for quick evaluation or equate coefficients for more complex cases.
Partial fractions are essential for integration (ln-type integrals) and for the method of differences.

Exam Tip: Always check that your fraction is proper before attempting partial fractions — if the degree of the numerator is $\geq$ the degree of the denominator, divide first. After finding $A, B, C, \ldots$ , always verify your answer by recombining the fractions and checking they give the original expression. The cover-up rule is fastest for distinct linear factors but does not work directly for repeated factors.

A-Level Deep Dive: Partial Fractions in Depth

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section B (Algebra and Functions), Year 2 content covers decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerators constant or linear) (refer to the official specification document for exact wording). Although decomposition is taught as an algebraic technique, AQA examines it in three downstream contexts: integration of rational functions (Pure Section H — partial fractions are the standard route to $\int \dfrac{p(x)}{q(x)}\,dx$ ), binomial general expansion (Section B again — expanding $(1 - 2x)^{-1}$ after splitting), and first-order ODEs in Year 2 mechanics modelling. The 7357 formula booklet does not list the cover-up rule or any decomposition templates — these must be memorised.

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks): Express $f(x) = \dfrac{2x^3 + 5x^2 + 4x + 3}{(x + 1)^2(x - 2)}$ in partial fractions.

Solution with mark scheme:

Step 1 — degree check. The numerator is degree 3; the denominator expands to degree 3 ( $(x+1)^2(x-2) = x^3 - 3x - 2$ after expansion). Since $\deg(\text{num}) \geq \deg(\text{denom})$ , the fraction is improper and we must polynomial-divide first.

M1 — recognising the improper fraction and starting polynomial division. Candidates who skip straight to a partial-fraction template here forfeit every mark — the template won't fit.

Step 2 — polynomial long division. Divide $2x^3 + 5x^2 + 4x + 3$ by $x^3 - 3x - 2$ :

$2x^3 + 5x^2 + 4x + 3 = 2 \cdot (x^3 - 3x - 2) + (5x^2 + 10x + 7)$

So $f(x) = 2 + \dfrac{5x^2 + 10x + 7}{(x + 1)^2(x - 2)}$ .

A1 — correct quotient $2$ and proper remainder $5x^2 + 10x + 7$ .

Step 3 — set up the proper-fraction template. Repeated linear factor $(x+1)^2$ contributes two terms; distinct linear factor $(x-2)$ contributes one:

$\dfrac{5x^2 + 10x + 7}{(x + 1)^2(x - 2)} \equiv \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{x - 2}$

M1 — correct template with both $A/(x+1)$ and $B/(x+1)^2$ for the repeated factor. Forgetting the second term is the single most common 7357 partial-fractions error.

Step 4 — clear denominators. Multiply through:

$5x^2 + 10x + 7 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)^2$

Step 5 — cover-up at $x = -1$ : $5 - 10 + 7 = 2 = B(-3)$ , so $B = -\tfrac{2}{3}$ .

M1 — applying cover-up at the repeated root to extract $B$ directly.

Step 6 — cover-up at $x = 2$ : $20 + 20 + 7 = 47 = C(9)$ , so $C = \tfrac{47}{9}$ .

A1 — $C$ correct.

Step 7 — find $A$ by comparing $x^2$ coefficients. LHS: $5$ . RHS: $A + C = A + \tfrac{47}{9}$ . So $A = 5 - \tfrac{47}{9} = \tfrac{45 - 47}{9} = -\tfrac{2}{9}$ .

A1 — $A$ correct, with sign carefully tracked.

Step 8 — assemble.

$f(x) = 2 - \dfrac{2}{9(x + 1)} - \dfrac{2}{3(x + 1)^2} + \dfrac{47}{9(x - 2)}$

A1 — final assembled form, polynomial part included.

Total: 8 marks (M3 A4, plus 1 method-mark for the template).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): Express $\dfrac{3x + 5}{(x - 1)(x^2 + 4)}$ in partial fractions, where the second denominator is irreducible over the reals.

Mark scheme decomposition by AO:

M1 (AO1.1a) — correct template $\dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 4}$ , recognising the irreducible quadratic requires a linear numerator.
M1 (AO1.1b) — clearing denominators: $3x + 5 = A(x^2 + 4) + (Bx + C)(x - 1)$ .
M1 (AO1.1b) — cover-up at $x = 1$ : $8 = 5A$ , so $A = \tfrac{8}{5}$ .
A1 (AO1.1b) — comparing $x^2$ : $0 = A + B$ , so $B = -\tfrac{8}{5}$ .
A1 (AO1.1b) — comparing constant: $5 = 4A - C$ , so $C = 4 \cdot \tfrac{8}{5} - 5 = \tfrac{32 - 25}{5} = \tfrac{7}{5}$ .
A1 (AO2.5) — final assembled form: $\dfrac{8}{5(x - 1)} + \dfrac{-8x + 7}{5(x^2 + 4)}$ .

Total: 6 marks split AO1 = 5, AO2 = 1. AQA reserves the AO2 mark for assembling and presenting the form cleanly — typically extracting the common factor of $\tfrac{1}{5}$ or matching the requested presentation.

Synoptic links

Connects to:

Integration of rational functions (Pure Section H): $\int \dfrac{p(x)}{q(x)}\,dx$ is intractable in raw form but trivial after partial-fraction decomposition. Each $\dfrac{A}{x - a}$ integrates to $A \ln|x - a|$ , each $\dfrac{B}{(x - a)^2}$ integrates to $-\dfrac{B}{x - a}$ , and each $\dfrac{Bx + C}{x^2 + k^2}$ splits into a $\ln$ piece and an $\arctan$ piece. Partial fractions is the gateway technique to integration in Year 2.
Binomial general expansion (Section B): expanding $\dfrac{1}{(1 - 2x)(1 + x)}$ as a Maclaurin series requires splitting first into $\dfrac{A}{1 - 2x} + \dfrac{B}{1 + x}$ , then expanding each piece using $(1 + u)^{-1} = 1 - u + u^2 - \ldots$ . Without partial fractions the binomial step is impossible.
Algebraic fractions (Section B): simplifying, adding and subtracting algebraic fractions is the reverse operation — starting from $\dfrac{A}{x + 1} + \dfrac{B}{x - 2}$ and combining over a common denominator. Fluency in both directions is essential.
First-order ODEs (Pure Section M, Year 2): separable equations like $\dfrac{dy}{dx} = \dfrac{y(1 - y)}{x}$ produce $\int \dfrac{1}{y(1 - y)}\,dy$ on the left, which requires partial-fraction splitting before integration. The logistic model is the canonical example.
Method of differences and telescoping series: identities like $\dfrac{1}{n(n + 1)} = \dfrac{1}{n} - \dfrac{1}{n + 1}$ are partial-fraction decompositions in disguise, used to evaluate $\sum_{n = 1}^{N} \dfrac{1}{n(n + 1)}$ exactly.

Mark-scheme literacy

Partial-fractions questions on 7357 split AO marks heavily toward AO1:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	70–80%	Setting up the correct template, clearing denominators, applying cover-up, comparing coefficients
AO2 (reasoning / interpretation)	15–25%	Recognising improper fractions, justifying the irreducible-quadratic numerator form, presenting in the requested form
AO3 (problem-solving)	0–10%	Open-ended modelling rare in isolation; appears synoptically in integration / ODE contexts

Partial Fractions in Depth

Partial Fractions in Depth

Why Partial Fractions?

Case 1: Distinct Linear Factors

Method: Cover-Up Rule (Heaviside)

Case 2: Repeated Linear Factors

Case 3: Improper Fractions

Applications

Application to Integration

Application to Series (Method of Differences)

Synoptic Links

Summary

A-Level Deep Dive: Partial Fractions in Depth

Spec mapping

Specimen Question (Modelled on the AQA Paper Format)

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

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