Partial fractions (HL)

Mathematics - Analysis and Approaches · IB Diploma Programme (DP) - SL & HL · 5 min read · Updated: 7 Sept 2026

Introduction to Partial Fractions

In your previous math studies, you learned how to take two separate fractions and add them together by finding a common denominator. For example, you know how to turn \( \frac{2}{x-1} + \frac{3}{x+2} \) into a single, more complex fraction.

Partial Fractions is simply that process in reverse! We start with one "big" fraction and break it down into its simpler "partial" components. This is a vital skill in the Higher Level (HL) course, especially when you reach Topic 5: Calculus, where breaking down fractions makes them much easier to integrate.

The Scope of the Syllabus

According to the AHL 1.11 syllabus requirements, you only need to master a specific type of partial fraction:

The denominator will have a maximum of two distinct linear terms (e.g., \( (x+a)(x+b) \)).
The numerator's degree will be less than the denominator's degree. This means if the bottom is a quadratic (\( x^2 \)), the top will be a linear expression (\( x \)) or just a constant number.

Note: You do not need to worry about "improper" fractions (where the top is bigger) or "repeated" factors for this specific sub-topic!

The General Setup

When you see a fraction with two distinct linear factors in the denominator, you can write it as the sum of two simpler fractions with unknown constants, \( A \) and \( B \).

The identity looks like this:

\( \frac{px + q}{(ax + b)(cx + d)} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} \)

Our goal is to find the specific numerical values for \( A \) and \( B \) that make this identity true for all values of \( x \).

Step-by-Step Guide: How to Decompose a Fraction

Don't worry if this seems abstract; the process is very logical. Let’s use the example: \( \frac{5x - 7}{(x - 1)(x - 2)} \)

Step 1: Set up the identity

Write the fraction as the sum of two parts with \( A \) and \( B \):

\( \frac{5x - 7}{(x - 1)(x - 2)} \equiv \frac{A}{x - 1} + \frac{B}{x - 2} \)

Step 2: Multiply by the common denominator

To get rid of the fractions, multiply every single term by the entire denominator \( (x - 1)(x - 2) \). This leaves you with a straight line of algebra:

\( 5x - 7 = A(x - 2) + B(x - 1) \)

Step 3: Solve for A and B

There are two popular ways to do this. We will focus on the Substitution Method because it is usually the fastest and easiest for these types of problems.

To find B, eliminate A:
Pick a value for \( x \) that makes the term with \( A \) become zero. Since we have \( A(x - 2) \), let’s choose \( x = 2 \):
\( 5(2) - 7 = A(2 - 2) + B(2 - 1) \)
\( 10 - 7 = 0 + B(1) \)
\( 3 = B \)

To find A, eliminate B:
Pick a value for \( x \) that makes the term with \( B \) become zero. Since we have \( B(x - 1) \), let’s choose \( x = 1 \):
\( 5(1) - 7 = A(1 - 2) + B(1 - 1) \)
\( 5 - 7 = A(-1) + 0 \)
\( -2 = -A \)
\( 2 = A \)

Step 4: Write the final answer

Substitute your values back into the original setup:
\( \frac{5x - 7}{(x - 1)(x - 2)} = \frac{2}{x - 1} + \frac{3}{x - 2} \)

Alternative Method: Equating Coefficients

Sometimes, students prefer comparing the coefficients. Using the same equation \( 5x - 7 = A(x - 2) + B(x - 1) \), you can expand the right side:

\( 5x - 7 = Ax - 2A + Bx - B \)
\( 5x - 7 = (A + B)x + (-2A - B) \)

Now, compare the two sides:

The \( x \) terms must match: \( 5 = A + B \)
The constant terms must match: \( -7 = -2A - B \)

You can solve these simultaneous equations (which you studied in previous chapters) to get the same result: \( A = 2 \) and \( B = 3 \).

Quick Review Box

1. Form: \( \frac{numerator}{(linear)(linear)} = \frac{A}{linear} + \frac{B}{linear} \)
2. Multiply: Get rid of the denominators.
3. Substitute: Use the roots of the denominator (the values of \( x \) that make the brackets zero) to find \( A \) and \( B \) quickly.

Common Mistakes to Avoid

Sign Errors: Be very careful with negative signs when multiplying or substituting. For example, if your factor is \( (x+3) \), you should substitute \( x = -3 \).
Mixing up A and B: Make sure you put the correct value over the correct denominator at the end.
Incomplete answers: Always write out the final decomposed expression; don't just stop once you find \( A \) and \( B \).

Why are we doing this? (The "Big Picture")

In Topic 5: Calculus (AHL 5.15), you will be asked to integrate functions like \( \int \frac{5x-7}{x^2-3x+2} dx \). This looks impossible at first! However, once you use partial fractions to turn it into \( \int (\frac{2}{x-1} + \frac{3}{x-2}) dx \), you can easily solve it using natural logarithms (\( \ln \)). Partial fractions take a "locked" algebraic expression and provide the "key" to open it for calculus.

Key Takeaway Summary

Partial Fractions is an algebraic technique used to split a rational function into a sum of simpler fractions. For your HL syllabus, you only need to handle cases with two distinct linear factors in the denominator where the numerator has a lower degree. Use the substitution method (choosing \( x \) values that zero out terms) to find the unknown constants quickly and accurately.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the purpose of partial fraction decomposition?

Partial fraction decomposition is the process of breaking down a complex rational expression into a sum of simpler "partial" fractions.

What are the specific constraints for partial fractions in the IB DP HL syllabus (AHL 1.11)?

The denominator must have a maximum of two distinct linear terms, and the degree of the numerator must be less than the degree of the denominator.

What is the general identity used to set up a partial fraction decomposition for two distinct linear factors?

\( \frac{px + q}{(ax + b)(cx + d)} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} \)

What is the first step in solving for the unknown constants \( A \) and \( B \)?

Multiply the entire equation by the common denominator to eliminate all fractions and create a linear algebraic equation.

How does the "Substitution Method" work for finding constants in partial fractions?

Choose values of \( x \) that make one of the terms (either the one with \( A \) or the one with \( B \)) equal to zero, allowing you to solve for the remaining constant.

What is the "Equating Coefficients" method for partial fractions?

Expand the algebraic expression and create a system of simultaneous equations by matching the coefficients of like terms (e.g., the \( x \) terms and the constants) from both sides.

Why is partial fraction decomposition essential for AHL Topic 5: Calculus?

It allows for the integration of complex rational functions by splitting them into simpler terms that can be integrated as natural logarithms (\( \ln \)).

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