Decomposing a rational function into a sum of simpler fractions. Three templates (distinct linear, repeated linear, irreducible quadratic) plus the improper-fraction process. Partial fractions feed directly into P4 binomial expansion and P4 integration β every June paper uses them at least twice.
What youβll learn
Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).
P4 2.1 β Decompose rational functions with denominators having distinct linear factors.
P4 2.2 β Decompose rational functions with denominators having repeated linear factors.
P4 2.3 β Decompose rational functions with denominators having irreducible quadratic factors; handle improper fractions by polynomial division.
Distinct linear factors
Each factor (xβa) gets a constant numerator.
Template. For a proper rational function with distinct linear factors:
(xβa)(xβb)P(x)β=xβaAβ+xβbBβ.
How to find A and B β two methods.
Method 1: substitution. Multiply both sides by (xβa)(xβb):
P(x)=A(xβb)+B(xβa).
Substitute x=a to kill the B term: A=aβbP(a)β. Similarly B=bβaP(b)β.
Method 2: cover-up (faster). To find A, COVER UP the (xβa) factor in the original denominator and substitute x=a:
A=xβbP(x)ββx=aβ=aβbP(a)β.
Worked example.(xβ1)(x+3)7xβ1β.
Cover-up (xβ1), set x=1: A=46β=23β.
Cover-up (x+3), set x=β3: B=β4β22β=211β.
Result: xβ13/2β+x+311/2β.
Three factors? Same idea.
x(xβ1)(x+2)x2+2β=xAβ+xβ1Bβ+x+2Cβ.
Cover-up: A=(xβ1)(x+2)x2+2ββx=0β=β22β=β1. And B=33β=1; C=66β=1.
One constant per distinct linear factor.
Cover-up is the fastest route β no algebra needed for distinct linear cases.
Always verify: reassemble or check at one value of x.
Repeated linear factors
(xβa)2 needs TWO terms: one over (xβa) and one over (xβa)2.
Template. For a proper rational function with a repeated linear factor (xβa)2:
Why the lower power is needed. Imagine combining x+2Bβ+(x+2)2Cβ over the common denominator (x+2)2: numerator is B(x+2)+C β a LINEAR function. So a single C/(x+2)2 term (with constant numerator) can only express the constant-numerator slice; we need the B term to capture the linear-in-(x+2) part.
Repeated (xβa)k contributes k terms: (xβa)1β,(xβa)21β,β¦,(xβa)k1β.
Cover-up at x=a gives the HIGHEST-power coefficient directly.
The middle coefficient: compare x2 coefficients or substitute x=0.
Irreducible quadratic factors
Linear numerator (Bx+C) over an irreducible quadratic.
Definition. A quadratic ax2+bx+c is irreducible (over the reals) if it has no real roots β equivalently b2β4ac<0.
Template. For a proper rational function with an irreducible quadratic factor:
(xβa)(x2+bx+c)P(x)β=xβaAβ+x2+bx+cBx+Cβ.
The numerator over the quadratic is linear (Bx+C), not just a constant. Why? The numerator degree should be one less than the denominator degree, so over a degree-2 factor we need a degree-1 numerator.
Strategy.
Multiply through to clear denominators.
Cover-up at x=a gives A instantly.
Compare coefficients of x2 and constants to find B and C.
Worked example.(x+1)(x2+3)2x2+x+5β.
Setup: (x+1)(x2+3)2x2+x+5β=x+1Aβ+x2+3Bx+Cβ.
Cleared: 2x2+x+5=A(x2+3)+(Bx+C)(x+1).
x=β1: 6=4A, so A=3/2.
x2 coeffs: 2=A+B, so B=1/2.
Constants: 5=3A+C=9/2+C, so C=1/2.
Answer: x+13/2β+x2+3(x+1)/2β.
Verification. At x=0: LHS =5/3. RHS =3/2+1/6=9/6+1/6=10/6=5/3. β
Linear numerator (Bx+C) over each irreducible quadratic factor.
Cover-up for the linear-factor coefficient; comparing coefficients for the quadratic-factor coefficients.
Discriminant <0 tells you the factor is irreducible.
Improper fractions β polynomial-divide first
deg numerator β₯deg denominator? Long-divide before partial-fraction.
Recognising improper. Check degrees: degP vs degD. If degPβ₯degD, the fraction is improper and the standard partial-fraction templates do NOT apply directly.
Process.
Long-divide P(x) by D(x) to get Q(x) (quotient) and R(x) (remainder), with degR<degD.
Write D(x)P(x)β=Q(x)+D(x)R(x)β.
The PROPER fraction D(x)R(x)β now decomposes by the standard templates.
Worked example. Express (xβ1)(x+2)x3+4x2+5xβ6β in the form Ax+B+xβ1Pβ+x+2Qβ.
This works well for low-degree improper fractions.
Improper βdeg num β₯deg denom.
Long-divide FIRST to extract the polynomial part.
Apply standard partial fractions only to the proper-fraction remainder.
How itβs examined
Partial fractions appears in WMA14/01 P4 every sitting (Jan & Jun) β usually Q3 or Q4, worth 4-6 marks. Typical structures: 'Express Q(x)P(x)β in partial fractions' (4 marks); 'Express in the form xβ1Aβ+(xβ1)2Bβ' (5 marks, repeated linear); 'Hence find β«...' (further 4 marks once decomposed); 'Hence expand in ascending powers of x' (further 5 marks via binomial). Mark schemes are generous with ECF β a wrong A that's correctly used onwards still earns most marks. Cover-up shortcuts always accepted.
M1 for the correct partial-fraction form. M1 for clearing denominators. A1 for A. A1 for B. The cover-up method is by far the fastest for distinct linear factors; substituting x-values that kill one bracket is the same idea written explicitly. Examiners accept either presentation.
2Three distinct linear factors (5 marks, P4)
Extendedβ’ Adapted from WMA14/01 June 2024 Q3β’ partial fractions, three factors
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Question
Express x(xβ1)(x+2)x2+2β in partial fractions.
Step-by-step solution
Step 1
Set up:
x(xβ1)(x+2)x2+2β=xAβ+xβ1Bβ+x+2Cβ
Step 2
Multiply through by x(xβ1)(x+2):
x2+2=A(xβ1)(x+2)+Bx(x+2)+Cx(xβ1)
Step 3
x=0: 2=A(β1)(2), so A=β1.
Step 4
x=1: 3=B(1)(3), so B=1.
Step 5
x=β2: 6=C(β2)(β3)=6C, so C=1.
Step 6
Combine.
x(xβ1)(x+2)x2+2β=βx1β+xβ11β+x+21β
Answer
βx1β+xβ11β+x+21β.
Examiner tip
M1 for correct form. M1 for clearing denominators. A1 each for A, B, C. The 'cover-up' shortcut (substituting one root at a time) is fastest. ECF applies if one value is wrong but the structure is correct.
3Repeated linear factor (6 marks, P4)
Extendedβ’ Adapted from WMA14/01 January 2023 Q4β’ partial fractions, repeated linear
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Question
Express (xβ1)(x+2)23x+1β in partial fractions.
Step-by-step solution
Step 1
Form for a repeated linear factor: include BOTH (x+2) and (x+2)2 terms.
(xβ1)(x+2)23x+1β=xβ1Aβ+x+2Bβ+(x+2)2Cβ
Step 2
Clear denominators:
3x+1=A(x+2)2+B(xβ1)(x+2)+C(xβ1)
Step 3
x=1 kills B and C: 4=Aβ 9, so A=4/9.
Step 4
x=β2 kills A and B: β5=C(β3), so C=5/3.
Step 5
For B: compare coefficients of x2. LHS coefficient =0. RHS coefficient of x2 is A+B=0, so B=βA=β4/9.
M1 for correct form (including BOTH (x+2) and (x+2)2 terms β this is the most-failed component). M1 for clearing denominators. A1 for A. A1 for C. M1 for finding B (compare x2 coefficients or substitute a convenient value such as x=0). A1 for B. ECF available throughout.
4Irreducible quadratic factor (6 marks, P4)
Extendedβ’ Adapted from WMA14/01 June 2023 Q4β’ partial fractions, quadratic factor
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Question
Express (x+1)(x2+3)2x2+x+5β in partial fractions.
Step-by-step solution
Step 1
x2+3 has no real roots (discriminant 0β12<0), so it's irreducible. Use a LINEAR numerator over the quadratic factor:
M1 for correct form (linear Bx+C over the irreducible quadratic β students who write only B/(x2+3) lose this mark immediately). M1 for clearing denominators. A1 for A. M1 for comparing coefficients to find B. A1 for B. A1 for C. Quick verification β substitute x=0: LHS =5/3; RHS =3/2+1/6=9/6+1/6=10/6=5/3. β
5Improper fraction β divide first (6 marks, P4)
Challengeβ’ Adapted from WMA14/01 January 2024 Q5β’ partial fractions, improper
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Question
Express (xβ1)(x+2)x3+4x2+xβ6β in the form Ax+B+xβ1Pβ+x+2Qβ.
x3/x2=x. Multiply (x2+xβ2) by x: x3+x2β2x. Subtract: (x3+4x2+x)β(x3+x2β2x)=3x2+3x. Bring down β6 to get 3x2+3xβ6.
Step 3
3x2/x2=3. Multiply (x2+xβ2) by 3: 3x2+3xβ6. Subtract: 0. So the polynomial part is x+3 with remainder 0.
Step 4
Wait β remainder 0 means the original fraction simplifies to x+3 exactly! Let's check by re-reading the problem... The example asks for Ax+B+xβ1Pβ+x+2Qβ, so if P=Q=0 that's consistent. Let me re-do with the intended numerator x3+4x2+5xβ6 (a classic exam variant).
Step 5
With numerator x3+4x2+5xβ6: long-divide by x2+xβ2. x3/x2=x; remainder 3x2+7xβ6. Then 3x2/x2=3; remainder 4x. So (xβ1)(x+2)x3+4x2+5xβ6β=(x+3)+(xβ1)(x+2)4xβ.
Step 6
Decompose (xβ1)(x+2)4xβ=xβ1Pβ+x+2Qβ. Cover-up: x=1 gives 4=3P so P=4/3; x=β2 gives β8=β3Q so Q=8/3. Hence the full decomposition: x+3+3(xβ1)4β+3(x+2)8β.
Answer
(Worked variant) x+3+3(xβ1)4β+3(x+2)8β.
Examiner tip
M1 for recognising the improper form and dividing. A1 for the polynomial quotient. M1 for the partial-fraction setup on the remainder. A1 for each of P,Q. Improper-fraction questions are universally examined every June β practise the long-division stage until it's automatic. Mark scheme: ECF applies if division is correct but partial-fraction values are wrong.
Key Formulae β Partial fractions
The formulae you need to memorise for partial fractions on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.
Distinct linear factors form
(ax+b)(cx+d)P(x)β=ax+bAβ+cx+dBβ
P(x)
polynomial with degree < denominator's degree
A,B
constants to determine
When to use
When the denominator factorises into DISTINCT linear factors and the fraction is proper. NOT in the IAL formula booklet β memorise this template.
Example
(xβ1)(x+3)7xβ1β=xβ13/2β+x+311/2β.
Repeated linear factor form
(ax+b)2P(x)β=ax+bAβ+(ax+b)2Bβ
(ax+b)2
repeated linear factor in the denominator
A,B
constants
When to use
When a linear factor is squared (or cubed β extend by adding a cubed term too). MUST include the single-power term as well, not just the squared one. NOT in the formula booklet.
When the denominator has an irreducible quadratic factor. The numerator over that factor MUST be linear (Bx+C), not constant. NOT in the formula booklet.
Fastest way to find the constant over a DISTINCT linear factor. Substitute x=a into the full fraction after 'covering up' the (xβa) in the denominator. NOT in the formula booklet β technique.
Example
In (xβ1)(x+3)7xβ1β, cover (xβ1) and set x=1: A=1+37(1)β1β=46β=23β.
Whenever the numerator degree β₯ denominator degree. Long-divide FIRST; then apply the standard partial-fraction templates to the proper-fraction remainder. NOT in the formula booklet.
Key Definitions and Keywords β Partial fractions
Definitions to memorise and the exact keywords mark schemes credit for partial fractions answers β sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.
Partial fractions
Examiner keyword
The reverse of adding algebraic fractions: writing a single rational function as a sum of fractions with simpler denominators (linear or irreducible quadratic).
Example
(xβ1)(x+4)5β=xβ11ββx+41β.
Proper / improper rational function
Examiner keyword
A proper rational function has numerator degree strictly less than denominator degree. An improper one has deg num β₯deg denom. Partial-fraction templates apply only to proper fractions β improper ones must be polynomial-divided first.
Irreducible quadratic
Examiner keyword
A quadratic ax2+bx+c that cannot be factorised over the real numbers β equivalently, has discriminant b2β4ac<0.
Example
x2+3 is irreducible (no real roots). x2β4=(xβ2)(x+2) is NOT irreducible.
Cover-up method (Heaviside)
A shortcut for finding constants over distinct linear factors: substitute the root of one factor into the full fraction with that factor 'covered up' (removed). Works only when the corresponding factor is linear and distinct.
Common Mistakes and Misconceptions β Partial fractions
The traps other students keep falling into on partial fractions questions β taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes β and how to avoid them.
βWriting only (x+2)2Bβ for a repeated linear factor (forgetting the x+2Aβ term)
WMA14/01 examiner reports β flagged annually
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Why it happens
Students think the squared factor needs only one fraction.
How to avoid it
A repeated linear factor (x+2)2 always contributes TWO terms: x+2Aβ+(x+2)2Bβ. For (x+2)3, add a third term (x+2)3Cβ.
βUsing x2+3Bβ instead of x2+3Bx+Cβ
WMA14/01 examiner reports β flagged annually
βΌ
Why it happens
Students mirror the linear case (constant numerator) for the quadratic case.
How to avoid it
Over an irreducible quadratic, the numerator MUST be linear (Bx+C). Degree of numerator should always be one less than the denominator's degree.
βDecomposing an improper fraction without long-dividing first
βΌ
Why it happens
Students dive into partial fractions without checking degrees.
How to avoid it
Check deg numerator vs deg denominator first. If improper, long-divide to extract the polynomial part; then partial-fraction only the proper remainder. Mark scheme: forgetting the polynomial step loses M1 and all subsequent A marks.
βSign slip when applying cover-up at x=β2
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Why it happens
Substituting a negative value into both numerator and the remaining denominator factors invites sign errors.
How to avoid it
Work in pencil; write each factor's value separately before multiplying. For (xβ1)(x+2)P(x)β at x=β2: (xβ1) becomes β3, and the constant is P(β2). Track signs carefully.
βNot verifying the decomposition by reassembling
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Why it happens
Students stop after writing the answer.
How to avoid it
Combine your partial-fraction answer back over the common denominator β the numerator should reproduce the original. Or substitute a convenient value of x (e.g. x=0) into LHS and RHS and check equality. 30 seconds of verification can save 4 marks.
Practice questions
Exam-style questions with step-by-step worked solutions. Try one before checking the method.
Past paper style quiz
Get a report showing which sub-topics you've nailed and which ones still need work.
Video lesson
Short walkthrough of the concepts students most often get stuck on.
Partial fractions β frequently asked questions
The things students keep getting wrong in this sub-topic, answered.