For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: decomposition with distinct linear, repeated linear, and irreducible quadratic factors within the K341 denominator limits forms the main route.
School-sensitive extension: improper fractions, more repeated factors, and calculus applications may vary by school.
2027 national comparison: K341 Topic A4.4 specifies partial fractions with denominators no more complicated than the named linear, repeated-linear, and linear-times-quadratic cases.
Check your school: confirm whether polynomial division is expected before decomposition and which denominator families are assessed.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 18) Partial Fractions cover? A: Decompose rational expressions with distinct, repeated, and irreducible quadratic factors in IP AMaths.
Partial fractions simplify rational functions for integration and algebraic manipulation.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Partial fractions split one rational expression into simpler pieces.
Use it as a working check: Factor the denominator first, choose the correct template, then solve for constants by substitution or coefficient comparison.
Then go one layer deeper: Example: if the denominator is (x + 4)(x - 1), write A/(x + 4) + B/(x - 1), multiply through, then use x = -4 and x = 1.
Choosing the partial fraction template
The denominator decides the template. Factor it fully before writing any constants.
Denominator pattern
Template move
Best way to solve constants
Distinct linear factors, such as (x+4)(x−1)
Put one constant over each linear factor.
Substitute the root that makes each factor zero.
Repeated linear factor, such as x2(x+3)
Include every power: xA+x2B
Substitute convenient roots first, then compare coefficients.
Irreducible quadratic factor, such as x2+4
Use a linear numerator Bx+C over the quadratic.
Compare coefficients after multiplying through.
Numerator degree is at least denominator degree
Divide first to make a proper fraction.
Decompose only the remainder fraction.
Common trap: do not put just one term over a repeated factor. For (x−a)2, both x−aA and (x−a)2B are needed.
Proper-fraction checkpoint
Before choosing a template, compare the degree of the numerator with the degree of the denominator.
Degree check
First action
Why
Numerator degree is lower
Decompose directly.
The fraction is already proper.
Numerator degree is equal
Divide first.
Partial fractions should be applied only to the proper remainder.
Numerator degree is higher
Divide first.
The quotient is a polynomial part that must stay outside the partial fractions.
Worked check:
x+1x2+3x+5=x+2+x+13.
Only x+13 is the proper fraction part. The quotient x+2 is part of the final answer, not a constant to hide inside A.
Common trap: do not start with x+1A when the original fraction is improper. That loses the polynomial part of the expression.
Constant-solving checkpoint
After multiplying through by the full denominator, choose the fastest way to find each constant instead of expanding immediately.
Situation after clearing denominators
First move
Why it works
Common trap
A distinct linear factor gives root x=a
Substitute x=a.
Every term with (x−a) becomes zero, so one constant is isolated.
Substituting before clearing denominators, which can create division by zero.
A repeated factor such as (x−a)2 appears
Substitute x=a first.
The highest-power term usually survives and gives one constant quickly.
Expecting the same substitution to find every constant in the repeated block.
Constants remain after useful substitutions
Compare coefficients of powers of x.
The identity must match for all values of x, so matching coefficients gives the missing equations.
Comparing coefficients before expanding and collecting like terms cleanly.
The numerator degree is too high
Divide first, then decompose the proper remainder.
Partial fractions apply to the proper fraction part.
Trying to hide the quotient inside the constants.
Misconception check: a substitution root is a shortcut after the denominator has been cleared; it is not saying the original fraction is defined at that root.
Repeated-factor slot checkpoint
For a repeated linear factor, write one partial-fraction slot for every power of that factor. Start with the first power and build up to the repeated power.
Denominator factor
Required slots
Why
(x−a)
x−aA
One linear factor needs one constant numerator.
(x−a)2
x−aA+(x−a)2B
(x−a)3
x−aA+(x−a)2B+(x−a)3C
Worked check: for x2(x+3)2x+1, the repeated factor is x2. The setup must include
xA+x2B+x+3C.
Misconception check: do not jump straight to (x−a)2A for a repeated factor. The missing x−aB term may be the part needed to match the numerator after clearing denominators.
Irreducible-quadratic numerator checkpoint
When a denominator contains an irreducible quadratic, the numerator above that quadratic must be linear. Use Bx+C, not just a constant.
Denominator factor
Correct numerator slot
Why
Common trap
x2+4
Bx+C
A quadratic factor needs enough freedom to match both the x term and constant term after clearing denominators.
Writing only B over the quadratic.
x2+px+q that cannot be factorised over real numbers
Bx+C
Do not force real linear factors when there are none.
Splitting it into fake factors such as (x+a)(x+b)
Linear factor multiplied by irreducible quadratic
linearA+quadraticBx+C
Worked check: for (x−2)(x2+3)5x+1, start with
x−2A+x2+3Bx+C.
Misconception check: x2+3 is not factorised into real linear factors, so substitution roots cannot isolate B and C. Clear denominators and compare coefficients.
1 Templates
Distinct linear factors: (x−a)(x−b)P(x)=x−aA+x−bB.
Multiply both sides by denominator: 7x+5=A(x−1)+B(x+4).
Substitute x=1: 12=5B⟹B=512
Substitute x=−4: −23=−5A⟹A=523
Therefore x2+3x−47x+5=5(x+4)23+5(x−1)12
3 Worked example - Repeated factor
Decompose x2(x+3)2x+1.
Form: x2(x+3)2x+1=xA+x2B+x+3C.
Multiply through: 2x+1=Ax(x+3)+B(x+3)+Cx2
Substitute x=0: 1=3B → B=31
Substitute x=−3: −5=C×9⟹C=−95
Compare coefficients for x2: left side has zero; right side gives A+C=0 so A=−C=95
Final form: x2(x+3)2x+1=9x5+3x21−9(x+3)5
4 Worked example - Linear factor with irreducible quadratic
Decompose
(x+1)(x2+4)3x2+5x+2
Assume:
(x+1)(x2+4)3x2+5x+2=x+1A+x2+4Bx+C
Multiply through:
3x2+5x+2=A(x2+4)+(Bx+C)(x+1) 3. Expand the right side and collect like terms: (A+B)x2+(B+C)x+(4A+C). 4. Match coefficients with the left side: A+B=3,B+C=5,4A+C=2. 5. Solve the system: from A+B=3,B=3−A. Using B+C=5 gives C=2+A. 6. Substitute into 4A+C=2 to get 4A+(2+A)=2⟹5A=0⟹A=0. 7. Hence B=3 and C=2.
Answer:
(x+1)(x2+4)3x2+5x+2=x2+43x+2.
5 Practice Quiz
Make sure decomposition templates, coefficient solving, and integration-ready forms are consistent before combining with calculus.
6 Try this
Decompose (x−1)2(x+2)5x2+7x−3 and note the coefficients clearly.
, then add the other factors.
Both powers are needed to make the identity flexible enough.
Omitting a lower-power slot can make the numerator impossible to match.
.
The linear factor gets a constant numerator; the quadratic gets a linear numerator.
