IP AMaths Notes (Upper Sec, Year 3-4): 18) Partial Fractions

Study guide08 Nov 2025, 00:00 ZUpdated 14 Oct 2026

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Marcus Pang·Managing Director (Maths)

Decompose rational expressions with distinct, repeated, and irreducible quadratic factors in IP AMaths.

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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.
Exam-track route: use the separate O-Level and SEC G3 Additional Mathematics notes for K341 topic ownership.

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.

New to the Integrated Programme? Start with What is IP? | Browse all free IP 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 $x^2(x + 3)$	Include every power: $\frac{A}{x} + \frac{B}{x^2}$ , then add the other factors.	Substitute convenient roots first, then compare coefficients.
Irreducible quadratic factor, such as $x^2 + 4$	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 $\frac{A}{x - a}$ and $\frac{B}{(x - a)^2}$ 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:

$\frac{x^2 + 3x + 5}{x + 1} = x + 2 + \frac{3}{x + 1}.$

Only $\frac{3}{x + 1}$ 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 $\frac{A}{x + 1}$ 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
$(x - a)$	$\dfrac{A}{x - a}$	One linear factor needs one constant numerator.
$(x - a)^2$	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$	Both powers are needed to make the identity flexible enough.
$(x - a)^3$	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2} + \dfrac{C}{(x - a)^3}$	Omitting a lower-power slot can make the numerator impossible to match.

Worked check: for $\dfrac{2x + 1}{x^2(x + 3)}$ , the repeated factor is $x^2$ . The setup must include

$\dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x + 3}.$

Misconception check: do not jump straight to $\dfrac{A}{(x - a)^2}$ for a repeated factor. The missing $\dfrac{B}{x - a}$ 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 has degree at most one; either coefficient may be zero. Use $Bx+C$ , not just a constant.

Denominator factor	Correct numerator slot	Why	Common trap
$x^2 + 4$	$Bx + C$	A quadratic factor needs enough freedom to match both the $x$ term and constant term after clearing denominators.	Writing only $B$ over the quadratic.
$x^2 + px + q$ 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	$\dfrac{A}{\text{linear}} + \dfrac{Bx+C}{\text{quadratic}}$	The linear factor gets a constant numerator; the quadratic gets a linear numerator.	Using $Ax+B$ over every denominator part.

Worked check: for $\dfrac{5x+1}{(x-2)(x^2+3)}$ , start with

$\dfrac{A}{x-2} + \dfrac{Bx+C}{x^2+3}.$

Misconception check: $x^2+3$ is not factorised into real linear factors, so substitution roots cannot isolate $B$ and $C$ . Clear denominators and compare coefficients.

1 Templates

These templates assume a proper fraction: the numerator has lower degree than the denominator. The two distinct linear roots must differ. Each identity holds on the original domain, excluding zeros of the denominator. An irreducible quadratic is taken over the real numbers, and its numerator has degree at most one.

Distinct linear factors: $\dfrac{P(x)}{(x - a)(x - b)} = \dfrac{A}{x - a} + \dfrac{B}{x - b}$ .
Repeated: $\dfrac{P(x)}{(x - a)^2} = \dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$ .
Irreducible quadratic: $\dfrac{P(x)}{(x^{2} + px + q)} = \dfrac{Ax + B}{x^{2} + px + q}$ .

2 Worked example - Distinct factors

Decompose $\dfrac{7x + 5}{x^{2} + 3x - 4}$ .

Factor denominator: $x^{2} + 3x - 4 = (x + 4)(x - 1)$ .
Set $\dfrac{7x + 5}{(x + 4)(x - 1)} = \dfrac{A}{x + 4} + \dfrac{B}{x - 1}$ .
Multiply both sides by denominator: $7x + 5 = A(x - 1) + B(x + 4)$ .
Substitute $x = 1$ : $12 = 5B \implies B = \dfrac{12}{5}$ .
Substitute $x = -4$ : $-23 = -5A \implies A = \dfrac{23}{5}$ .
Therefore $\dfrac{7x + 5}{x^{2} + 3x - 4} = \dfrac{23}{5(x + 4)} + \dfrac{12}{5(x - 1)}$ .

3 Worked example - Repeated factor

Decompose $\dfrac{2x + 1}{x^{2}(x + 3)}$ .

Form: $\dfrac{2x + 1}{x^{2}(x + 3)} = \dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x + 3}$ .
Multiply through: $2x + 1 = A x(x + 3) + B(x + 3) + C x^{2}$ .
Substitute $x = 0$ : $1 = 3B$ → $B = \dfrac{1}{3}$ .
Substitute $x = -3$ : $-5 = C \times 9 \implies C = -\dfrac{5}{9}$ .
Compare coefficients for $x^{2}$ : left side has zero; right side gives $A + C = 0$ so $A = -C = \dfrac{5}{9}$ .
Final form: $\dfrac{2x + 1}{x^{2}(x + 3)} = \dfrac{5}{9x} + \dfrac{1}{3x^{2}} - \dfrac{5}{9(x + 3)}$ .

4 Worked example - Linear factor with irreducible quadratic

Decompose

$\frac{3x^2 + 5x + 2}{(x + 1)(x^2 + 4)}$

1. Assume:

$\frac{3x^2 + 5x + 2}{(x + 1)(x^2 + 4)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 4}$

2. Multiply through:

$3x^2 + 5x + 2 = A(x^2 + 4) + (Bx + C)(x + 1)$

3. Expand the right side and collect like terms: $(A + B)x^2 + (B + C)x + (4A + C)$ . 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 $4A + (2 + A) = 2 \implies 5A = 0 \implies A = 0$ . 7. Hence $B = 3$ and $C = 2$ .

Answer: $\frac{3x^2 + 5x + 2}{(x + 1)(x^2 + 4)} = \frac{3x + 2}{x^2 + 4}.$

The original expression remains undefined at x = -1 after cancellation; the simplified form agrees with it only on the original domain.

5 Practice Quiz

Make sure decomposition templates, coefficient solving, and integration-ready forms are consistent before combining with calculus.

6 Try this

Decompose $\dfrac{5x^{2} + 7x - 3}{(x - 1)^2(x + 2)}$ and note the coefficients clearly.

Sources

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