If you have ever stared at a complex algebraic fraction in your A-Math paper and wondered how on earth you are supposed to simplify it, partial fractions is the answer. This topic sits neatly within the Additional Mathematics (A-Math) syllabus and is one of those skills that, once mastered, opens the door to more advanced work in calculus and integration later on. The good news is that partial fraction decomposition follows a set of clear, predictable rules — and once you understand the logic behind each case, the process becomes surprisingly straightforward.

In this guide, we break down all four cases of partial fractions tested in A-Math: distinct linear factors, repeated linear factors, irreducible quadratic factors, and improper fractions that require long division first. Whether you are encountering this topic for the first time or brushing up before your O-Level examinations, this step-by-step walkthrough will give you the clarity and confidence to handle any partial fraction question that comes your way.

What Are Partial Fractions?

Partial fractions are the simpler fractions you obtain when you break apart a more complex rational expression. Think of it as the reverse of combining fractions over a common denominator. When you add 2/(x + 1) and 3/(x − 2), you get a single algebraic fraction. Partial fraction decomposition works in the opposite direction — starting from that combined fraction and splitting it back into its simpler components. This process is also called partial fraction expansion.

This skill is not just an abstract algebraic exercise. It has real applications in calculus, where integrating a complex rational function becomes much easier once it has been decomposed into simpler partial fractions. At the A-Math level, however, your primary task is simply to express a given algebraic fraction as a sum of its partial fractions — and to do that correctly, you need to recognise which of the four cases applies.

Proper vs. Improper Algebraic Fractions

Before you can decompose any algebraic fraction, you must determine whether it is proper or improper. This distinction is critical because partial fraction decomposition only works directly on proper fractions.

The degree of a polynomial is its highest power of x. An algebraic fraction f(x)/g(x) is classified as follows:

• Proper fraction: The degree of the numerator f(x) is less than the degree of the denominator g(x). For example, (3x + 1) / [(x + 2)(x − 1)] is proper because the numerator has degree 1 and the denominator has degree 2.
• Improper fraction: The degree of the numerator f(x) is equal to or greater than the degree of the denominator g(x). For example, (x³ + 2x) / [(x − 1)(x + 3)] is improper because the numerator has degree 3 and the denominator has degree 2.

If your fraction is improper, you must carry out polynomial long division first to convert it into the form of a polynomial plus a proper fraction. Only then can you apply partial fraction decomposition to the proper remainder. This step is so important in A-Math that it constitutes its own case — Case 4 — which we cover in detail below.

General Steps for Partial Fraction Decomposition

Regardless of which case you are dealing with, the overall process for decomposing a fraction into partial fractions follows the same logical sequence. Getting comfortable with these steps will make every partial fractions question feel more manageable.

1. Check if the fraction is proper or improper — Compare the degree of the numerator with the degree of the denominator. If the fraction is improper, perform polynomial long division first and work with the proper remainder.
2. Fully factorise the denominator — Make sure the denominator is broken down into all its factors. This determines which case (or combination of cases) applies. Do not forget useful identities like difference of two squares from Secondary 2.
3. Set up the correct partial fraction form — Based on the type of factors in the denominator, write out the appropriate partial fraction template with unknown constants (typically labelled A, B, C, etc.) in the numerators.
4. Multiply both sides by the denominator (LCM) — Clearing all the denominators gives you an algebraic identity, which is true for all values of x.
5. Solve for the unknown constants — Use substitution of strategic x-values, comparison of coefficients, or a combination of both.
6. Write the final answer — Substitute the values of A, B, C back into your partial fraction template and simplify.

Now let us look at exactly how each of the four cases works.

Case 1: Distinct Linear Factors

This is the most straightforward case and the one most students encounter first. The denominator contains two or more distinct linear factors — meaning linear expressions of the form (ax + b) that are all different from one another and each appears only once. For each distinct linear factor in the denominator, you assign a separate partial fraction with a constant in the numerator.

The standard form looks like this:

f(x) / [(ax + b)(cx + d)] = A / (ax + b) + B / (cx + d)

Each partial fraction must itself be a proper fraction, which is why each numerator holds a constant (not a polynomial) when the denominator factor is linear.

Worked Example — Case 1

Express (5x + 1) / [(x + 1)(x − 2)] in partial fractions.

Since both factors are distinct and linear, set up: (5x + 1) / [(x + 1)(x − 2)] = A / (x + 1) + B / (x − 2)

Multiply both sides by (x + 1)(x − 2): 5x + 1 = A(x − 2) + B(x + 1)

Let x = 2: 11 = 3B, so B = 11/3
Let x = −1: −4 = −3A, so A = 4/3

Therefore: (5x + 1) / [(x + 1)(x − 2)] = (4/3)/(x + 1) + (11/3)/(x − 2)

Case 2: Repeated Linear Factors

A repeated linear factor occurs when the same linear expression appears more than once in the denominator — written as (ax + b)². This case catches many students off guard because the instinct is to write only one partial fraction for that factor. The correct approach, however, is to include a separate partial fraction for each power of the repeated factor, from the first power up to the highest.

The standard form is:

f(x) / [(ax + b)²] = A / (ax + b) + B / (ax + b)²

If you have both a distinct factor and a repeated factor together, such as (cx + d)(ax + b)², the decomposition becomes: A / (cx + d) + B / (ax + b) + C / (ax + b)²

Worked Example — Case 2

Express (3x + 5) / [(x + 1)(x − 2)²] in partial fractions.

Set up: (3x + 5) / [(x + 1)(x − 2)²] = A / (x + 1) + B / (x − 2) + C / (x − 2)²

Multiply both sides by (x + 1)(x − 2)²: 3x + 5 = A(x − 2)² + B(x + 1)(x − 2) + C(x + 1)

Let x = 2: 11 = 3C, so C = 11/3
Let x = −1: 2 = 9A, so A = 2/9
Compare coefficients of x²: 0 = A + B, so B = −2/9

A key point to remember: when a linear factor repeats, you must include both A/(ax + b) and B/(ax + b)² in your setup. Omitting either term is one of the most common errors in A-Math partial fraction questions.

Case 3: Irreducible Quadratic Factors

An irreducible quadratic factor is a quadratic expression in the denominator that cannot be factorised any further over the real numbers. In A-Math, this typically appears in the form x² + c², where c is a constant. You can confirm a quadratic is irreducible by checking its discriminant: if b² − 4ac < 0, the quadratic has no real roots and cannot be factorised.

Because the denominator factor has degree 2, the corresponding partial fraction must have a linear expression (Bx + C) in the numerator, not just a constant. Using only a constant would fail to account for all possible numerator forms that keep the partial fraction proper.

The standard form is:

f(x) / [(ax + b)(x² + c²)] = A / (ax + b) + (Bx + C) / (x² + c²)

Worked Example — Case 3

Express (x² − x + 10) / [(x + 1)(x² + 3)] in partial fractions.

Check: numerator degree (2) < denominator degree (3) — this is a proper fraction. The factor (x² + 3) is irreducible since its discriminant is 0 − 12 = −12 < 0.

Set up: (x² − x + 10) / [(x + 1)(x² + 3)] = A / (x + 1) + (Bx + C) / (x² + 3)

Multiply both sides by (x + 1)(x² + 3): x² − x + 10 = A(x² + 3) + (Bx + C)(x + 1)

Let x = −1: 1 + 1 + 10 = 4A, so 12 = 4A, giving A = 3
Let x = 0: 10 = 3A + C = 9 + C, so C = 1
Compare coefficients of x²: 1 = A + B = 3 + B, so B = −2

Therefore: (x² − x + 10) / [(x + 1)(x² + 3)] = 3/(x + 1) + (−2x + 1)/(x² + 3)

Case 4: Improper Fractions (Long Division First)

This case applies whenever the degree of the numerator is equal to or greater than the degree of the denominator. Because partial fraction decomposition only works on proper fractions, you must first use polynomial long division to rewrite the expression in the form:

Quotient + Remainder / Denominator

The quotient is a polynomial (which you leave as is), and the remainder over the denominator is now a proper fraction that you can decompose using Cases 1, 2, or 3 as appropriate.

Worked Example — Case 4

Express (x³ + 3x² − 2x + 1) / [(x + 3)(x − 1)] in partial fractions.

Step 1 — Check degree: Numerator degree is 3; denominator degree is 2. This is an improper fraction.

Step 2 — Expand the denominator: (x + 3)(x − 1) = x² + 2x − 3

Step 3 — Perform long division of (x³ + 3x² − 2x + 1) by (x² + 2x − 3). The result gives a quotient of (x + 1) and a remainder of (−x + 4), so:

(x³ + 3x² − 2x + 1) / (x² + 2x − 3) = (x + 1) + (−x + 4) / [(x + 3)(x − 1)]

Step 4 — Decompose the proper remainder using Case 1 (distinct linear factors):

(−x + 4) / [(x + 3)(x − 1)] = A / (x + 3) + B / (x − 1)

−x + 4 = A(x − 1) + B(x + 3)
Let x = 1: 3 = 4B, so B = 3/4
Let x = −3: 7 = −4A, so A = −7/4

Final answer: (x³ + 3x² − 2x + 1) / [(x + 3)(x − 1)] = (x + 1) − (7/4)/(x + 3) + (3/4)/(x − 1)

Finding the Unknown Constants: Two Methods

Once you have cleared the denominators and obtained an algebraic identity, you need to find the values of all the unknown constants. There are two main methods, and knowing when to use each one will save you significant time in the examination.

Method 1: Substituting Strategic Values of x

The fastest approach is to substitute values of x that cause one or more terms to disappear, isolating a single unknown. For distinct linear factors, substitute the root of each factor in turn. For example, if one factor is (x − 3), let x = 3 to zero out that entire term. This method works particularly well for Cases 1 and 2 and is the go-to approach when you can find values that cleanly eliminate terms.

Method 2: Comparing Coefficients

When substitution alone cannot determine all the constants — especially in Case 3 involving an irreducible quadratic — you need to expand both sides of the identity and match the coefficients of corresponding powers of x. For instance, compare the coefficients of x², x¹, and the constant term separately on the left and right-hand sides to set up a system of equations. This method is always valid and is essential when no substitution will isolate the remaining unknowns cleanly.

In practice, the smartest strategy is to use substitution where possible to find some constants first, then use coefficient comparison to find the rest. Combining both methods in a single question is perfectly acceptable and often the most efficient path.

Common Mistakes to Avoid

Partial fractions is a topic where students consistently lose marks in predictable ways. Being aware of these pitfalls before your exam is just as valuable as knowing the correct methods.

• Forgetting to check for improper fractions: Jumping straight into decomposition without checking the degrees of the numerator and denominator is the single most costly error. Always compare degrees first.
• Incorrect numerator for an irreducible quadratic: Writing a constant (e.g., C) instead of a linear expression (Cx + D) over a quadratic factor in the denominator is a very common Case 3 error.
• Missing terms for repeated factors: For a factor like (x − 2)², you need both A/(x − 2) and B/(x − 2)². Omitting the A/(x − 2) term means your decomposition is incomplete.
• Incomplete factorisation of the denominator: If the denominator is not fully factorised before you set up the partial fractions, your decomposition will be wrong from the start. Always check whether any remaining quadratic can be factorised further.
• Arithmetic errors when solving for constants: Take your time substituting values and expanding brackets. A small sign error or arithmetic slip will make the entire answer wrong.

Exam Tips for Partial Fractions

Partial fractions questions in A-Math papers are usually worth between 4 and 7 marks, meaning there are multiple marks on offer — and multiple ways to earn partial credit even if you make one error. Here are some targeted exam strategies to maximise your score:

• Always state your setup clearly: Write out the full partial fraction template (with A, B, C placeholders) before solving. Markers award method marks for a correctly structured setup, even if the final constants contain an arithmetic error.
• Verify your answer by adding the partial fractions back: If time permits, recombine your partial fractions over a common denominator and check that you recover the original expression. This is a reliable self-check.
• For Case 3, try both methods together: Use substitution to find constants associated with the linear factor first, then compare coefficients to find the constants in the quadratic numerator. This combination is almost always faster than coefficient comparison alone.
• Do not skip the long division step: When the fraction is improper, show your long division working clearly. Marks are awarded for the division itself, not just the final decomposed form.
• Check your discriminant for quadratic factors: Before treating a quadratic as irreducible, quickly compute b² − 4ac. If the discriminant is non-negative, the quadratic can be factorised and you should apply Case 1 instead.

Students who do well in partial fractions questions tend to be systematic rather than fast. Slow down at the setup stage, choose the right case, and the algebra will follow naturally. This is exactly the kind of methodical thinking we nurture at EduFirst — through our small-group secondary tuition classes, students get the focused practice and personalised feedback they need to internalise these processes confidently.

Putting It All Together

Partial fractions is one of those A-Math topics that rewards careful preparation. Once you understand the four cases — distinct linear factors, repeated linear factors, irreducible quadratic factors, and improper fractions requiring long division — the method becomes a structured, repeatable process rather than a mystery. The key is always to start by checking whether the fraction is proper, factorise the denominator fully, set up the correct partial fraction template for each case, and then solve for the unknown constants using substitution, coefficient comparison, or both.

With consistent practice, you will begin to recognise which case applies at a glance, dramatically reducing the time you spend on these questions in the actual exam. Partial fractions also lays the groundwork for integration later in the A-Math syllabus, so investing time in it now pays dividends across the entire course. If you are looking for structured guidance and worked examples across the full Additional Mathematics syllabus, explore EduFirst’s secondary tuition programmes or try our flexible e-lessons for learning at your own pace.