IB Maths AA HL Topic 1 — Number & Algebra HL only Paper 1 & 2 ~9 min read

Partial Fractions

Partial fractions is the reverse of adding fractions together. If you can compute 2x − 1 + 3x + 4 and get a single fraction, partial fractions just runs that backwards: take the messy single fraction and split it back into clean pieces. It’s HL-only, and once the procedure clicks it’s mechanical. Where it really earns its keep is later — for integration, the binomial expansion of rational functions, and Maclaurin series, all of which need rational expressions split first.

📘 What you need to know

Partial fractions reverses the addition of algebraic fractions.
The shape of the split depends on the denominator: distinct linear factors, repeated linear factors, or numerator-degree ≥ denominator-degree.
For each distinct linear factor like (x − a), include one term Ax − a.
For a repeated linear factor like (x − a)², include two terms: Ax − a + B(x − a)².
If the numerator’s degree ≥ the denominator’s, do polynomial division first.
Find the unknown constants by substituting roots and/or comparing coefficients.
The IB exam usually tells you the form — your job is to find the constants.

What partial fractions actually do

Forwards is easy: combine two fractions into one. Backwards is what we want.

Forwards (already familiar) 2x − 1 + 3x + 4 = 2(x + 4) + 3(x − 1)(x − 1)(x + 4) = 5x + 5(x − 1)(x + 4)

Backwards (partial fractions) 5x + 5(x − 1)(x + 4) = 2x − 1 + 3x + 4 ✗ method, not a formula — nothing to look up in the booklet

🧠 Why bother?

Single rational fractions are usually impossible to integrate directly. But once split, each piece looks like 1x − a — and that integrates to ln|x − a|. Partial fractions converts a hard problem into a sum of easy ones. Same trick works for series expansions.

The three cases you’ll meet at HL

The denominator’s structure tells you the shape of the answer. Recognise the case first; then plug in.

CASE 1

Two distinct linear factors

stuff
──────────
(x − a)(x − b)

↓

A
────
x − a + B
────
x − b

CASE 2

A repeated linear factor

stuff
──────
(x − a)²

↓

A
────
x − a + B
──────
(x − a)²

CASE 3

Numerator degree ≥ denominator

ax + b
────────
cx + d

↓

A + B
────
cx + d

🤔 Why does a repeated factor need two terms?

Because B(x − a)² alone has a constant on top — but the numerator that produced (x − a)² in the denominator could have been linear. Adding Ax − a + B(x − a)² over a common denominator gives A(x − a) + B(x − a)² — and that linear top is exactly what’s needed to capture the original fully.

The method — finding the constants

🧭 Recipe — Case 1 & Case 2

Factorise the denominator (if not already done).
Set up the partial fraction form with unknown constants A, B, … — match the case shape.
Multiply both sides by the original denominator to clear the fractions.
Find the constants using either:
• substitution — pick x-values that make brackets zero (the roots), or
• comparing coefficients — match the coefficient of each power of x on both sides.
Write the final answer with constants in place.

Method A: Substituting roots (the “cover-up” trick)

Once you’ve cleared denominators, plug in the roots of each linear factor. This zeros out one of the unknowns and leaves a single equation in the other.

Why it works: setting x = a kills the (x − a) factor — so the term carrying the corresponding unknown vanishes, and you can solve for the other constants directly.

Method B: Comparing coefficients

Expand the right-hand side, then match the coefficient of x², x, and the constant term on each side. You get a system of simultaneous equations to solve.

For Cases 1 and 2, substitution is faster most of the time. For repeated factors, it gets you two of the three constants — you’ll usually need to compare coefficients (or substitute one extra value) to pin down the third. Mix the two methods freely; whatever’s quickest.

When the numerator is “too big” (Case 3)

If the degree of the numerator is the same as or higher than the denominator, the fraction is improper. You can’t split it directly — first do polynomial division to peel off a constant (or polynomial), then handle the leftover proper fraction normally.

Linear over linear ax + bcx + d = A + Bcx + d

Quick check: if the highest power of x on top is ≥ highest power on the bottom, it’s improper. Divide first.

Worked examples

WE 1

Two distinct linear factors — already factorised

Express 4x + 5(x − 1)(x + 2) in partial fractions.

Step 1: Set up the form 4x + 5(x − 1)(x + 2) ≡ Ax − 1 + Bx + 2 Step 2: Multiply both sides by (x − 1)(x + 2) 4x + 5 = A(x + 2) + B(x − 1) Step 3: Substitute the roots Let x = 1 : 4 + 5 = 3A → A = 3 Let x = −2 : −8 + 5 = −3B → B = 1 = 3x − 1 + 1x + 2

WE 2

Denominator needs factorising first

Express 4x − 1x² − 3x − 4 in partial fractions.

Step 1: Factorise the denominator x² − 3x − 4 = (x − 4)(x + 1) Step 2: Set up the form and clear denominators 4x − 1(x − 4)(x + 1) ≡ Ax − 4 + Bx + 1 4x − 1 = A(x + 1) + B(x − 4) Step 3: Substitute roots x = 4 : 16 − 1 = 5A → A = 3 x = −1 : −4 − 1 = −5B → B = 1 = 3x − 4 + 1x + 1

WE 3

Pure repeated factor

Express 5x − 3(x − 1)² in partial fractions.

Step 1: Set up — repeated factor needs two terms 5x − 3(x − 1)² ≡ Ax − 1 + B(x − 1)² Step 2: Multiply by (x − 1)² 5x − 3 = A(x − 1) + B Step 3: Substitute the root x = 1 5 − 3 = 0 + B → B = 2 Step 4: Compare coefficients of x left: 5 ; right: A → A = 5 = 5x − 1 + 2(x − 1)² substitution gave B, comparing coefficients gave A — both methods used together

WE 4

Linear and repeated linear together

Express 3x² − 5x + 1(x + 1)(x − 2)² in the form Ax + 1 + Bx − 2 + C(x − 2)².

Step 1: Multiply through by (x + 1)(x − 2)² 3x² − 5x + 1 = A(x − 2)² + B(x + 1)(x − 2) + C(x + 1) Step 2: Substitute the roots to peel off two unknowns x = −1 : 3 + 5 + 1 = 9A → A = 1 x = 2 : 12 − 10 + 1 = 3C → C = 1 Step 3: Compare x² coefficients to find B left: 3 ; right: A + B 3 = 1 + B → B = 2 = 1x + 1 + 2x − 2 + 1(x − 2)² substitute the two roots first — that’s free unknowns. Compare coefficients only for the leftover.

WE 5

Linear over linear (improper)

Express 3x + 5x + 1 in the form A + Bx + 1.

Step 1: Multiply through by (x + 1) 3x + 5 = A(x + 1) + B Step 2: Substitute the root x = −1 −3 + 5 = B → B = 2 Step 3: Compare coefficient of x left: 3 ; right: A → A = 3 = 3 + 2x + 1 same answer as polynomial division: 3x + 5 ÷ (x + 1) = 3 remainder 2

WE 6

Exam-style — form is given, find the constants

Given that 7x + 4x² + x − 6 ≡ Px + 3 + Qx − 2, find the values of P and Q.

Step 1: Factorise — confirms the form is correct x² + x − 6 = (x + 3)(x − 2) Step 2: Multiply through by (x + 3)(x − 2) 7x + 4 = P(x − 2) + Q(x + 3) Step 3: Substitute the two roots x = 2 : 18 = 5Q → Q = 185 x = −3 : −17 = −5P → P = 175 P = 175, Q = 185 fractional answers are fine — leave them as fractions in IB exams

💡 Top tips

Read the question carefully — most IB partial-fraction questions tell you the form. Match your set-up exactly to whatever they’ve written.
Always factorise the denominator first. If the denominator is a quadratic, factor it before deciding the case.
Substituting roots is the fastest method. Use it to grab as many constants as possible before reaching for compare-coefficients.
For repeated factors, you’ll need both methods. Substitute the root for the “deeper” constant, then compare coefficients for the rest.
Check by recombining. Quickly add your partial fractions back together — should reproduce the original numerator.
Improper fractions need division first. If degree of top ≥ degree of bottom, polynomial-divide before splitting.
Fractions in your final answer are fine. If A = 175, leave it like that — don’t decimalise.

⚠ Common mistakes

Using only one term for a repeated factor. (x − a)² in the denominator needs both Ax − a and B(x − a)², not just one.
Splitting an improper fraction directly. If the top’s degree ≥ the bottom’s, you must divide first or the partial-fraction form won’t work.
Forgetting to multiply through correctly. Each term on the right gets multiplied by the full original denominator — easy to drop a factor.
Sign errors with negatives. When substituting x = −2, that’s −2 (negative two), not 2. Use brackets.
Rounding too early. Keep constants as exact fractions; decimal approximations introduce errors when the answer is later checked or reused.
Not factorising first. Trying partial fractions on stuffx² + x − 6 without factorising the bottom — won’t get anywhere.
Mixing up A and B. If you label one fraction’s constant A, make sure the substitution and final answer both refer to the same fraction.

Partial fractions is one of those topics that looks intimidating the first time and feels routine the tenth. Drill the three cases until you can read the form straight off the denominator — that’s most of the speed gain right there. After that, you’re just plugging in roots.

Need help with Partial Fractions?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →