Partial Fractions — AHL 1.11 Partial fractions (two...

SECTION 1

Historical Context & Motivation

The idea of breaking a fraction into simpler parts has deep roots in algebra. Long before calculators existed, mathematicians needed ways to simplify complicated rational expressions so they could perform calculations by hand. The technique of partial fraction decomposition grew out of this practical need and became a cornerstone of both algebra and calculus. Today, partial fractions remain essential in the IB Mathematics curriculum because they unlock methods for integration and series that would otherwise be inaccessible.

1702

Leibniz and Johann Bernoulli

Gottfried Wilhelm Leibniz and Johann Bernoulli debated the integration of rational functions. Their correspondence highlighted the need for a systematic way to decompose fractions before integrating.

1740

Euler Formalises the Method

Leonhard Euler developed rigorous rules for splitting rational expressions into sums of simpler fractions. He applied partial fractions extensively in his work on infinite series and differential equations.

1800s

Standard Algebra Curriculum

Partial fractions became a standard topic in university mathematics across Europe. Textbooks began presenting the technique as a fundamental algebraic tool alongside polynomial division and factoring.

Present

IB & Modern Applications

Partial fractions appear in the IB Mathematics: Analysis and Approaches syllabus (AHL 1.11) and are used in engineering, signal processing, and computer science to simplify transfer functions and Laplace transforms.

The central question this topic answers is: given a fraction whose denominator factors into two distinct linear terms, how can we express it as the sum of two simpler fractions? Mastering this technique opens the door to more advanced work in integration and algebraic manipulation throughout the IB course.

SECTION 2

Core Principles & Definitions

Before diving into the mechanics of partial fractions, you need to be comfortable with a few foundational ideas. A rational expression is any fraction where both the numerator and denominator are polynomials — for example, (3x + 5) / (x² − x − 6). The goal of partial fraction decomposition is to reverse the process of adding fractions: instead of combining two simple fractions into one complex one, you take a complex fraction apart into simpler pieces.

Proper Fraction

A rational expression where the degree of the numerator is strictly less than the degree of the denominator. Partial fractions only apply directly to proper fractions.

Distinct Linear Factors

The denominator factors into terms of the form (ax + b) where no factor is repeated. For example, (x − 2)(x + 3) has two distinct linear factors.

Decomposition Form

For two distinct linear factors, the decomposition takes the form A/(x − r₁) + B/(x − r₂), where A and B are constants you need to find.

Identity Principle

The decomposition must hold for all values of x (except where the denominator is zero). This fact lets you set up equations to solve for A and B.

✦ KEY TAKEAWAY

Think of partial fractions like reversing a recipe. When you add 1/3 + 1/4, you combine them into 7/12. Partial fraction decomposition is like tasting the 7/12 mixture and figuring out the original 1/3 and 1/4 ingredients. You already know how to combine fractions — now you're learning to take them apart.

SECTION 3

Visual Explanation

The diagram below illustrates the core idea of partial fraction decomposition. On the left, you see a single complex fraction with a factored quadratic denominator. The arrow shows how this single fraction splits into two simpler fractions, each with one of the original linear factors as its denominator. The constants A and B in the numerators are determined by the algebraic methods you will learn in the next sections.

The diagram shows a rational expression with two distinct linear factors in its denominator being split into two simpler fractions. The constants A and B are found by substitution or by equating coefficients.

Notice how the original fraction on the left has a quadratic denominator (degree 2) and a linear numerator (degree 1), which makes it a proper fraction. The two resulting fractions on the right each have simpler linear denominators. This decomposition is only valid when the denominator factors into two distinct linear factors — that is, the factors must be different from each other. If they were the same (a repeated factor), a different decomposition form would be required.

SECTION 4

Mathematical Framework

Let's set up the algebraic framework step by step. Suppose you have a proper rational expression whose denominator can be factored into two distinct linear factors. The general form and its decomposition are shown below.

GENERAL DECOMPOSITION

(px + q) / ((x − a)(x − b)) = A / (x − a) + B / (x − b)

where p, q are known coefficients in the numerator, a, b are the roots of the denominator (a ≠ b), and A, B are the unknown constants to determine.

To find A and B, multiply both sides of the equation by the full denominator (x − a)(x − b). This clears all fractions and gives you a polynomial identity.

CLEARING THE DENOMINATOR

px + q = A(x − b) + B(x − a)

This identity holds for all values of x. This means you can substitute any convenient value of x to create simple equations for A and B.

Method 1: Substitution (Cover-Up Method)

The substitution method (sometimes called the cover-up method) is the fastest approach. Since the identity px + q = A(x − b) + B(x − a) is true for every x, choose x-values that make one term vanish.

FINDING A

Set x = a: pa + q = A(a − b) + 0 → A = (pa + q) / (a − b)

Substituting x = a makes the B-term disappear because (a − a) = 0, leaving a simple equation for A.

FINDING B

Set x = b: pb + q = 0 + B(b − a) → B = (pb + q) / (b − a)

Substituting x = b makes the A-term disappear, giving a direct equation for B.

Method 2: Equating Coefficients

An alternative approach is to expand the right side of px + q = A(x − b) + B(x − a) and then match coefficients of like powers of x. Expanding gives (A + B)x + (−Ab − Ba) = px + q. Equating the coefficient of x gives A + B = p, and equating the constant terms gives −Ab − Ba = q. You then solve this system of two equations for two unknowns. Both methods yield identical results, so use whichever feels more natural.

SECTION 5

Step-by-Step Process & Classification

Before performing partial fraction decomposition, you must confirm two conditions: the fraction is proper, and the denominator consists of distinct linear factors. The flowchart below walks you through the complete decision process from start to finish.

This flowchart guides you through the complete partial fractions process. Start at the top by checking that your fraction is proper, then factor the denominator, set up the template, clear denominators, and solve for the unknown constants.

💡 IB Exam Tip

In the IB exam, the question will usually tell you to "express in partial fractions" or the denominator will already be factored for you. However, you should still be ready to factor a quadratic denominator yourself using the techniques from AHL 1.9. Always double-check that a ≠ b (the factors are distinct) before applying this form.

SECTION 6

Worked Example

Let's work through a complete example from start to finish. We will decompose the fraction (5x + 1) / (x² + x − 6) into partial fractions.

Decompose (5x + 1) / (x² + x − 6)

Step 1 — Check the fraction is properThe numerator 5x + 1 has degree 1, and the denominator x² + x − 6 has degree 2. Since 1 < 2, this is a proper fraction, so we can proceed directly.

✓ Proper fraction confirmed

Step 2 — Factor the denominatorWe need two numbers that multiply to −6 and add to +1. Those numbers are +3 and −2, so: x² + x − 6 = (x + 3)(x − 2). These are two distinct linear factors with a = −3 and b = 2.

x² + x − 6 = (x + 3)(x − 2)

Step 3 — Write the decomposition templateFollowing the standard form, we write: (5x + 1) / ((x + 3)(x − 2)) = A / (x + 3) + B / (x − 2), where A and B are constants we need to find.

(5x + 1) / ((x + 3)(x − 2)) = A/(x + 3) + B/(x − 2)

Step 4 — Multiply both sides by (x + 3)(x − 2)Clearing the denominators gives the identity: 5x + 1 = A(x − 2) + B(x + 3). This equation must be true for all values of x.

5x + 1 = A(x − 2) + B(x + 3)

Step 5 — Substitute x = 2 to find BSetting x = 2 eliminates the A-term: 5(2) + 1 = A(2 − 2) + B(2 + 3) gives 11 = 0 + 5B, so B = 11/5.

B = 11/5

Step 6 — Substitute x = −3 to find ASetting x = −3 eliminates the B-term: 5(−3) + 1 = A(−3 − 2) + B(−3 + 3) gives −14 = −5A + 0, so A = 14/5.

A = 14/5

Step 7 — Write the final answerSubstituting A = 14/5 and B = 11/5 back into the template gives the completed decomposition.

(5x + 1) / ((x + 3)(x − 2)) = (14/5)/(x + 3) + (11/5)/(x − 2)

Step 8 — Verify (optional but recommended)To check, recombine: [14(x − 2) + 11(x + 3)] / [5(x + 3)(x − 2)] = [14x − 28 + 11x + 33] / [5(x + 3)(x − 2)] = (25x + 5) / [5(x + 3)(x − 2)] = (5x + 1) / [(x + 3)(x − 2)]. This matches the original — our decomposition is correct. ✓

✓ Verification complete

SECTION 7

Comparing Solution Methods

You now know two methods for finding the constants A and B: substitution and equating coefficients. Each method has its own strengths, and the table below compares them side by side. Understanding when to use each method will help you solve problems more efficiently in exams.

Comparison of the two main methods for finding partial fraction constants

Feature	Substitution Method	Equating Coefficients
Speed	Very fast — each substitution directly gives one constant	Slower — requires expanding, collecting terms, and solving a system
Ease of use	Simple arithmetic; great for two distinct linear factors	More algebraic steps; better when substitution doesn't eliminate terms neatly
Error risk	Low — fewer steps means fewer chances for mistakes	Moderate — sign errors in expansion are common
Applicability	Works perfectly for distinct linear factors	Works for any type of partial fraction (including repeated and quadratic factors)
IB recommendation	Preferred for AHL 1.11	Useful as a check or for more complex cases

⚡ WHEN IN DOUBT, SUBSTITUTE

For problems with two distinct linear factors — which is exactly what AHL 1.11 focuses on — the substitution method is almost always faster and less error-prone. Think of it like having a shortcut on a familiar road: the scenic route (equating coefficients) gets you there too, but the shortcut saves time. Save equating coefficients for when you encounter repeated or quadratic factors in later topics.

SECTION 8

Connection to Advanced Partial Fractions

The technique you've learned for two distinct linear factors is the simplest case of partial fraction decomposition. As you progress through IB Mathematics and into university-level calculus, you will encounter more complex scenarios. The table below previews these extensions to give you a roadmap of what's ahead.

Types of partial fraction decomposition and when they appear in the curriculum

Denominator Type	Decomposition Form	Where You'll See It
Two distinct linear factors (x − a)(x − b)	A/(x − a) + B/(x − b)	AHL 1.11 — This lesson!
Three or more distinct linear factors	A/(x − a) + B/(x − b) + C/(x − c) + ...	Extension of AHL 1.11; same method, more terms
Repeated linear factor (x − a)²	A/(x − a) + B/(x − a)²	University calculus; different template
Irreducible quadratic factor (x² + bx + c)	(Ax + B)/(x² + bx + c)	University calculus; numerator is linear, not constant

The good news is that the core logic remains the same in every case: set up a template, clear denominators, and solve for the unknowns. Mastering the two-factor case thoroughly gives you a strong foundation that extends naturally to all these more advanced scenarios. In the IB course specifically, partial fractions connect directly to integration of rational functions, where decomposing a fraction allows you to integrate each simpler piece using the natural logarithm.

🔗 Link to Integration

Once you can write f(x) = A/(x − a) + B/(x − b), you can integrate directly: ∫f(x)dx = A·ln|x − a| + B·ln|x − b| + C. This is one of the main reasons partial fractions are taught in the IB — they are a gateway to integrating rational functions.

SECTION 9

Practice Problems

Test your understanding with these five problems, arranged from conceptual understanding to critical thinking. Try each one on paper before revealing the answer.

PROBLEM 1 — CONCEPTUAL

Explain why the fraction (x² + 1) / ((x − 1)(x + 2)) cannot be decomposed into partial fractions in the form A/(x − 1) + B/(x + 2) without first performing an additional step. What is that step?

PROBLEM 2 — BASIC CALCULATION

Express (3x − 1) / ((x − 1)(x + 2)) in partial fractions.

PROBLEM 3 — INTERMEDIATE

Decompose (7x + 13) / (2x² + 7x + 3) into partial fractions. (Hint: start by factoring the denominator.)

PROBLEM 4 — APPLIED

In a biology model, the population growth rate is given by R(t) = 6 / ((t + 1)(t + 3)), where t is time in years. Express R(t) in partial fractions and use your result to describe what happens to each term as t increases.

PROBLEM 5 — CRITICAL THINKING

Prove that for any proper fraction (px + q) / ((x − a)(x − b)) with a ≠ b, the constants in the partial fraction decomposition always satisfy A + B = p. What does this tell you about the behaviour of the original fraction for large values of x?

SUMMARY

Lesson Summary

Partial fraction decomposition is the process of expressing a proper rational expression as a sum of simpler fractions. When the denominator factors into two distinct linear factors (x − a)(x − b), the decomposition takes the form A/(x − a) + B/(x − b). You find the constants A and B by clearing denominators and then either using the substitution method (setting x equal to each root) or by equating coefficients of like powers of x.

Remember the key prerequisites: confirm the fraction is proper (degree of numerator < degree of denominator) and that the factors are distinct. The substitution method is fastest for two distinct linear factors, and always verify your answer by recombining the partial fractions. This technique connects directly to integration of rational functions using natural logarithms, making it an essential tool in the IB Analysis and Approaches Higher Level course.