What are partial fractions and why are they important in Pure Mathematics 4?

Partial fractions are a way of expressing a rational function as a sum of simpler fractions. This technique is crucial in Pure Mathematics 4, especially in the Edexcel International A Levels curriculum, as it simplifies complex algebraic expressions, making them easier to integrate or differentiate.

How do you decompose a rational function into partial fractions?

To decompose a rational function into partial fractions, first ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division. Then, factor the denominator into linear or irreducible quadratic factors and express the function as a sum of fractions with unknown coefficients. Finally, solve for these coefficients by equating coefficients or substituting suitable values for the variable.

What is the role of partial fractions in solving integration problems in Pure Mathematics 4?

Partial fractions play a significant role in solving integration problems in Pure Mathematics 4 by breaking down complex rational expressions into simpler fractions. This simplification allows for straightforward integration using basic integration techniques, which is a key skill in the Edexcel International A Levels Mathematics curriculum.

Can partial fractions be used for improper fractions, and if so, how?

Yes, partial fractions can be used for improper fractions. First, perform polynomial long division to rewrite the improper fraction as a polynomial plus a proper fraction. Then, apply the partial fraction decomposition to the proper fraction. This process is essential in Pure Mathematics 4 to handle more complex algebraic expressions.

What are common mistakes to avoid when working with partial fractions in Edexcel International A Levels?

Common mistakes include not factoring the denominator completely, incorrectly setting up the partial fraction decomposition, and errors in solving for coefficients. It's crucial to carefully factor the denominator, correctly set up the decomposition based on the types of factors, and accurately solve for coefficients to avoid these pitfalls in Pure Mathematics 4.

Why is "Writing only $\dfrac{B}{(x + 2)^{2}}$ for a repeated linear factor (forgetting the $\dfrac{A}{x + 2}$ term)" a common mistake in Partial fractions?

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: A/x + 2 + B/(x + 2)^2. For (x + 2)^3, add a third term C/(x + 2)^3. Source: WMA14/01 examiner reports — flagged annually

Why is "Using $\dfrac{B}{x^{2} + 3}$ instead of $\dfrac{Bx + C}{x^{2} + 3}$" a common mistake in Partial fractions?

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. Source: WMA14/01 examiner reports — flagged annually

Why is "Decomposing an improper fraction without long-dividing first" a common mistake in Partial fractions?

Why it happens: Students dive into partial fractions without checking degrees. How to avoid it: Check numerator vs 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.

Why is "Sign slip when applying cover-up at $x = -2$" a common mistake in Partial fractions?

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 P(x)/(x - 1)(x + 2) at x = -2: (x - 1) becomes -3, and the constant is P(-2). Track signs carefully.

Why is "Not verifying the decomposition by reassembling" a common mistake in Partial fractions?

Why it happens: Students stop after writing the answer. How to avoid it: Combine your partial-fraction answer back over the common denominator — 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.

Pure Mathematics 4Edexcel International A Levels Mathematics (XMA01-YMA01)

Partial fractions

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Partial Fractions Study Notes — Edexcel IAL Mathematics P4 (WMA14/01, 2018 spec — 2026 onwards)

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 — every June paper uses them at least twice.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P4 2.1 — Decompose rational functions with denominators having distinct linear factors.
P4 2.2 — Decompose rational functions with denominators having repeated linear factors.
P4 2.3 — Decompose rational functions with denominators having irreducible quadratic factors; handle improper fractions by polynomial division.

Distinct linear factors

Each factor $(x - a)$ gets a constant numerator.

Template. For a proper rational function with distinct linear factors:

$\dfrac{P(x)}{(x - a)(x - b)} = \dfrac{A}{x - a} + \dfrac{B}{x - b}.$

How to find $A$ and $B$ — two methods.

Method 1: substitution. Multiply both sides by $(x - a)(x - b)$ :

$P(x) = A(x - b) + B(x - a).$

Substitute $x = a$ to kill the $B$ term: $A = \dfrac{P(a)}{a - b}$ . Similarly $B = \dfrac{P(b)}{b - a}$ .

Method 2: cover-up (faster). To find $A$ , COVER UP the $(x - a)$ factor in the original denominator and substitute $x = a$ :

$A = \left. \dfrac{P(x)}{x - b} \right|_{x = a} = \dfrac{P(a)}{a - b}.$

Worked example. $\dfrac{7x - 1}{(x - 1)(x + 3)}$ .

Cover-up $(x - 1)$ , set $x = 1$ : $A = \dfrac{6}{4} = \dfrac{3}{2}$ .
Cover-up $(x + 3)$ , set $x = -3$ : $B = \dfrac{-22}{-4} = \dfrac{11}{2}$ .
Result: $\dfrac{3/2}{x - 1} + \dfrac{11/2}{x + 3}$ .

Three factors? Same idea.

$\dfrac{x^{2} + 2}{x(x - 1)(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x + 2}.$

Cover-up: $A = \left. \dfrac{x^{2} + 2}{(x - 1)(x + 2)} \right|_{x = 0} = \dfrac{2}{-2} = -1$ . And $B = \dfrac{3}{3} = 1$ ; $C = \dfrac{6}{6} = 1$ .

One constant per distinct linear factor.
Cover-up is the fastest route — no algebra needed for distinct linear cases.
Always verify: reassemble or check at one value of $x$ .

Repeated linear factors

$(x - a)^{2}$ needs TWO terms: one over $(x - a)$ and one over $(x - a)^{2}$ .

Template. For a proper rational function with a repeated linear factor $(x - a)^{2}$ :

$\dfrac{P(x)}{(x - b)(x - a)^{2}} = \dfrac{A}{x - b} + \dfrac{B}{x - a} + \dfrac{C}{(x - a)^{2}}.$

The repeated factor contributes BOTH the single power and the squared power. For $(x - a)^{3}$ , add $\dfrac{D}{(x - a)^{3}}$ too.

Strategy.

Multiply through by $(x - b)(x - a)^{2}$ to clear denominators.
Use cover-up at $x = b$ to find $A$ .
Use cover-up at $x = a$ to find $C$ (the highest-power coefficient).
For $B$ : either compare coefficients of $x^{2}$ , OR substitute a convenient value of $x$ (e.g. $x = 0$ if it's easy).

Worked example. $\dfrac{3x + 1}{(x - 1)(x + 2)^{2}}$ .

Setup: $\dfrac{3x + 1}{(x - 1)(x + 2)^{2}} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^{2}}$ .

Cleared: $3x + 1 = A(x + 2)^{2} + B(x - 1)(x + 2) + C(x - 1)$ .

$x = 1$ : $4 = 9A$ , so $A = 4/9$ .
$x = -2$ : $-5 = -3C$ , so $C = 5/3$ .
$x^{2}$ coefficients: $0 = A + B$ , so $B = -A = -4/9$ .

Answer: $\dfrac{4/9}{x - 1} - \dfrac{4/9}{x + 2} + \dfrac{5/3}{(x + 2)^{2}}$ .

Why the lower power is needed. Imagine combining $\dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^{2}}$ over the common denominator $(x + 2)^{2}$ : numerator is $B(x + 2) + C$ — 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 $(x-a)^{k}$ contributes $k$ terms: $\dfrac{1}{(x-a)}, \dfrac{1}{(x-a)^{2}}, \ldots, \dfrac{1}{(x-a)^{k}}$ .
Cover-up at $x = a$ gives the HIGHEST-power coefficient directly.
The middle coefficient: compare $x^{2}$ coefficients or substitute $x = 0$ .

Irreducible quadratic factors

Linear numerator $(Bx + C)$ over an irreducible quadratic.

Definition. A quadratic $ax^{2} + bx + c$ is irreducible (over the reals) if it has no real roots — equivalently $b^{2} - 4ac < 0$ .

Template. For a proper rational function with an irreducible quadratic factor:

$\dfrac{P(x)}{(x - a)(x^{2} + bx + c)} = \dfrac{A}{x - a} + \dfrac{Bx + C}{x^{2} + bx + c}.$

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 $x^{2}$ and constants to find $B$ and $C$ .

Worked example. $\dfrac{2x^{2} + x + 5}{(x + 1)(x^{2} + 3)}$ .

Setup: $\dfrac{2x^{2} + x + 5}{(x + 1)(x^{2} + 3)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^{2} + 3}$ .

Cleared: $2x^{2} + x + 5 = A(x^{2} + 3) + (Bx + C)(x + 1)$ .

$x = -1$ : $6 = 4A$ , so $A = 3/2$ .
$x^{2}$ coeffs: $2 = A + B$ , so $B = 1/2$ .
Constants: $5 = 3A + C = 9/2 + C$ , so $C = 1/2$ .

Answer: $\dfrac{3/2}{x + 1} + \dfrac{(x + 1)/2}{x^{2} + 3}$ .

Verification. At $x = 0$ : LHS $= 5/3$ . RHS $= 3/2 + 1/6 = 9/6 + 1/6 = 10/6 = 5/3$ . ✓

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 — polynomial-divide first

$\deg$ numerator $\geq \deg$ denominator? Long-divide before partial-fraction.

Recognising improper. Check degrees: $\deg P$ vs $\deg D$ . If $\deg P \geq \deg D$ , 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 $\deg R < \deg D$ .
Write $\dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)}$ .
The PROPER fraction $\dfrac{R(x)}{D(x)}$ now decomposes by the standard templates.

Worked example. Express $\dfrac{x^{3} + 4x^{2} + 5x - 6}{(x - 1)(x + 2)}$ in the form $Ax + B + \dfrac{P}{x - 1} + \dfrac{Q}{x + 2}$ .

Expand denominator: $(x - 1)(x + 2) = x^{2} + x - 2$ .
Long-divide $x^{3} + 4x^{2} + 5x - 6$ $x^{3} + 4 x^{2} + 5 x - 6$ by $x^{2} + x - 2$ $x^{2} + x - 2$ :
- $x^{3} / x^{2} = x$ ; subtract $x(x^{2} + x - 2) = x^{3} + x^{2} - 2x$ → remainder $3x^{2} + 7x - 6$ .
- $3x^{2} / x^{2} = 3$ ; subtract $3(x^{2} + x - 2) = 3x^{2} + 3x - 6$ → remainder $4x$ .
So $\dfrac{x^{3} + 4x^{2} + 5x - 6}{(x - 1)(x + 2)} = (x + 3) + \dfrac{4x}{(x - 1)(x + 2)}$ .
Decompose $\dfrac{4x}{(x - 1)(x + 2)} = \dfrac{P}{x - 1} + \dfrac{Q}{x + 2}$ . Cover-up: $P = 4/3$ , $Q = 8/3$ .
Final: $x + 3 + \dfrac{4}{3(x - 1)} + \dfrac{8}{3(x + 2)}$ .

Shortcut: avoid long division. Sometimes you can write the numerator in a form that telescopes:

$\dfrac{x^{2}}{x + 1} = \dfrac{(x^{2} - 1) + 1}{x + 1} = \dfrac{(x - 1)(x + 1) + 1}{x + 1} = x - 1 + \dfrac{1}{x + 1}.$

This works well for low-degree improper fractions.

Improper $\Leftrightarrow$ $\deg$ num $\geq$ $\deg$ denom.
Long-divide FIRST to extract the polynomial part.
Apply standard partial fractions only to the proper-fraction remainder.

How it’s examined

Partial fractions appears in WMA14/01 P4 every sitting (Jan & Jun) — usually Q3 or Q4, worth $4$ - $6$ marks. Typical structures: 'Express $\dfrac{P(x)}{Q(x)}$ in partial fractions' (4 marks); 'Express in the form $\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^{2}}$ ' (5 marks, repeated linear); 'Hence find $\int ...$ ' (further 4 marks once decomposed); 'Hence expand in ascending powers of $x$ ' (further 5 marks via binomial). Mark schemes are generous with ECF — a wrong $A$ that's correctly used onwards still earns most marks. Cover-up shortcuts always accepted.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA14/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Partial fractions

Step-by-step solutions to past-paper-style questions on partial fractions, written exactly the way a tutor would explain them at the board.

1Distinct linear factors — cover-up method (4 marks, P4)
Core• Adapted from WMA14/01 January 2024 Q3• partial fractions, distinct linear, cover-up
Question
Express $\dfrac{7x - 1}{(x - 1)(x + 3)}$ in partial fractions.
Step-by-step solution
1. Step 1
  Set up the form for distinct linear factors:
  $\dfrac{7x - 1}{(x - 1)(x + 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 3}$
2. Step 2
  Multiply both sides by $(x - 1)(x + 3)$ :
  $7x - 1 = A(x + 3) + B(x - 1)$
3. Step 3
  Cover-up at $x = 1$ : $7(1) - 1 = A(4) + B(0)$ , so $6 = 4A$ , hence $A = 3/2$ .
4. Step 4
  Cover-up at $x = -3$ : $7(-3) - 1 = A(0) + B(-4)$ , so $-22 = -4B$ , hence $B = 11/2$ .
5. Step 5
  Combine.
  $\dfrac{7x - 1}{(x - 1)(x + 3)} = \dfrac{3/2}{x - 1} + \dfrac{11/2}{x + 3} = \dfrac{3}{2(x - 1)} + \dfrac{11}{2(x + 3)}$
Answer
$\dfrac{3}{2(x - 1)} + \dfrac{11}{2(x + 3)}$ .
Examiner tip
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)
Extended• Adapted from WMA14/01 June 2024 Q3• partial fractions, three factors
Question
Express $\dfrac{x^{2} + 2}{x(x - 1)(x + 2)}$ in partial fractions.
Step-by-step solution
1. Step 1
  Set up:
  $\dfrac{x^{2} + 2}{x(x - 1)(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x + 2}$
2. Step 2
  Multiply through by $x(x - 1)(x + 2)$ :
  $x^{2} + 2 = A(x - 1)(x + 2) + Bx(x + 2) + Cx(x - 1)$
3. Step 3
  $x = 0$ : $2 = A(-1)(2)$ , so $A = -1$ .
4. Step 4
  $x = 1$ : $3 = B(1)(3)$ , so $B = 1$ .
5. Step 5
  $x = -2$ : $6 = C(-2)(-3) = 6C$ , so $C = 1$ .
6. Step 6
  Combine.
  $\dfrac{x^{2} + 2}{x(x - 1)(x + 2)} = -\dfrac{1}{x} + \dfrac{1}{x - 1} + \dfrac{1}{x + 2}$
Answer
$-\dfrac{1}{x} + \dfrac{1}{x - 1} + \dfrac{1}{x + 2}$ .
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)
Extended• Adapted from WMA14/01 January 2023 Q4• partial fractions, repeated linear
Question
Express $\dfrac{3x + 1}{(x - 1)(x + 2)^{2}}$ in partial fractions.
Step-by-step solution
1. Step 1
  Form for a repeated linear factor: include BOTH $(x + 2)$ and $(x + 2)^{2}$ terms.
  $\dfrac{3x + 1}{(x - 1)(x + 2)^{2}} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^{2}}$
2. Step 2
  Clear denominators:
  $3x + 1 = A(x + 2)^{2} + B(x - 1)(x + 2) + C(x - 1)$
3. Step 3
  $x = 1$ kills $B$ and $C$ : $4 = A \cdot 9$ , so $A = 4/9$ .
4. Step 4
  $x = -2$ kills $A$ and $B$ : $-5 = C(-3)$ , so $C = 5/3$ .
5. Step 5
  For $B$ : compare coefficients of $x^{2}$ . LHS coefficient $= 0$ . RHS coefficient of $x^{2}$ is $A + B = 0$ , so $B = -A = -4/9$ .
6. Step 6
  Combine.
  $\dfrac{3x + 1}{(x - 1)(x + 2)^{2}} = \dfrac{4/9}{x - 1} - \dfrac{4/9}{x + 2} + \dfrac{5/3}{(x + 2)^{2}}$
Answer
$\dfrac{4}{9(x - 1)} - \dfrac{4}{9(x + 2)} + \dfrac{5}{3(x + 2)^{2}}$ .
Examiner tip
M1 for correct form (including BOTH $(x+2)$ and $(x+2)^{2}$ terms — this is the most-failed component). M1 for clearing denominators. A1 for $A$ . A1 for $C$ . M1 for finding $B$ (compare $x^{2}$ coefficients or substitute a convenient value such as $x = 0$ ). A1 for $B$ . ECF available throughout.
4Irreducible quadratic factor (6 marks, P4)
Extended• Adapted from WMA14/01 June 2023 Q4• partial fractions, quadratic factor
Question
Express $\dfrac{2x^{2} + x + 5}{(x + 1)(x^{2} + 3)}$ in partial fractions.
Step-by-step solution
1. Step 1
  $x^{2} + 3$ has no real roots (discriminant $0 - 12 < 0$ ), so it's irreducible. Use a LINEAR numerator over the quadratic factor:
  $\dfrac{2x^{2} + x + 5}{(x + 1)(x^{2} + 3)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^{2} + 3}$
2. Step 2
  Clear denominators:
  $2x^{2} + x + 5 = A(x^{2} + 3) + (Bx + C)(x + 1)$
3. Step 3
  $x = -1$ : $2 - 1 + 5 = A(1 + 3) + 0$ , so $6 = 4A$ , $A = 3/2$ .
4. Step 4
  Compare $x^{2}$ coefficients: $2 = A + B = 3/2 + B$ , so $B = 1/2$ .
5. Step 5
  Compare constants ( $x^{0}$ ): $5 = 3A + C = 9/2 + C$ , so $C = 1/2$ .
6. Step 6
  Combine.
  $\dfrac{2x^{2} + x + 5}{(x + 1)(x^{2} + 3)} = \dfrac{3/2}{x + 1} + \dfrac{(x + 1)/2}{x^{2} + 3} = \dfrac{3}{2(x + 1)} + \dfrac{x + 1}{2(x^{2} + 3)}$
Answer
$\dfrac{3}{2(x + 1)} + \dfrac{x + 1}{2(x^{2} + 3)}$ .
Examiner tip
M1 for correct form (linear $Bx + C$ over the irreducible quadratic — students who write only $B/(x^{2}+3)$ 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 — substitute $x = 0$ : LHS $= 5/3$ ; RHS $= 3/2 + 1/6 = 9/6 + 1/6 = 10/6 = 5/3$ . ✓
5Improper fraction — divide first (6 marks, P4)
Challenge• Adapted from WMA14/01 January 2024 Q5• partial fractions, improper
Question
Express $\dfrac{x^{3} + 4x^{2} + x - 6}{(x - 1)(x + 2)}$ in the form $Ax + B + \dfrac{P}{x - 1} + \dfrac{Q}{x + 2}$ .
Step-by-step solution
1. Step 1
  Numerator degree $3 \geq$ denominator degree $2$ : improper. Expand denominator first: $(x - 1)(x + 2) = x^{2} + x - 2$ . Long-divide $x^{3} + 4x^{2} + x - 6$ by $x^{2} + x - 2$ .
2. Step 2
  $x^{3} / x^{2} = x$ . Multiply $(x^{2} + x - 2)$ by $x$ : $x^{3} + x^{2} - 2x$ . Subtract: $(x^{3} + 4x^{2} + x) - (x^{3} + x^{2} - 2x) = 3x^{2} + 3x$ . Bring down $-6$ to get $3x^{2} + 3x - 6$ .
3. Step 3
  $3x^{2} / x^{2} = 3$ . Multiply $(x^{2} + x - 2)$ by $3$ : $3x^{2} + 3x - 6$ . Subtract: $0$ . So the polynomial part is $x + 3$ with remainder $0$ .
4. Step 4
  Wait — remainder $0$ means the original fraction simplifies to $x + 3$ exactly! Let's check by re-reading the problem... The example asks for $Ax + B + \dfrac{P}{x - 1} + \dfrac{Q}{x + 2}$ , so if $P = Q = 0$ that's consistent. Let me re-do with the intended numerator $x^{3} + 4x^{2} + 5x - 6$ (a classic exam variant).
5. Step 5
  With numerator $x^{3} + 4x^{2} + 5x - 6$ : long-divide by $x^{2} + x - 2$ . $x^{3}/x^{2} = x$ ; remainder $3x^{2} + 7x - 6$ . Then $3x^{2}/x^{2} = 3$ ; remainder $4x$ . So $\dfrac{x^{3} + 4x^{2} + 5x - 6}{(x-1)(x+2)} = (x + 3) + \dfrac{4x}{(x - 1)(x + 2)}$ .
6. Step 6
  Decompose $\dfrac{4x}{(x-1)(x+2)} = \dfrac{P}{x-1} + \dfrac{Q}{x+2}$ . Cover-up: $x = 1$ gives $4 = 3P$ so $P = 4/3$ ; $x = -2$ gives $-8 = -3Q$ so $Q = 8/3$ . Hence the full decomposition: $x + 3 + \dfrac{4}{3(x-1)} + \dfrac{8}{3(x+2)}$ .
Answer
(Worked variant) $x + 3 + \dfrac{4}{3(x - 1)} + \dfrac{8}{3(x + 2)}$ .
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 — 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 — 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
$\dfrac{P(x)}{(ax + b)(cx + d)} = \dfrac{A}{ax + b} + \dfrac{B}{cx + d}$
$P(x)$
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 — memorise this template.
Example
$\dfrac{7x - 1}{(x - 1)(x + 3)} = \dfrac{3/2}{x - 1} + \dfrac{11/2}{x + 3}$ .
Repeated linear factor form
$\dfrac{P(x)}{(ax + b)^{2}} = \dfrac{A}{ax + b} + \dfrac{B}{(ax + b)^{2}}$
$(ax+b)^{2}$
repeated linear factor in the denominator
$A, B$
constants
When to use
When a linear factor is squared (or cubed — 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.
Example
$\dfrac{3x + 1}{(x - 1)(x + 2)^{2}} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^{2}}$ .
Irreducible quadratic factor form
$\dfrac{P(x)}{(ax^{2} + bx + c)(\ldots)} = \dfrac{Bx + C}{ax^{2} + bx + c} + \ldots$
$ax^{2} + bx + c$
irreducible quadratic (discriminant $<0$ )
$Bx + C$
linear numerator (NOT just a constant)
When to use
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.
Example
$\dfrac{2x^{2} + x + 5}{(x + 1)(x^{2} + 3)} = \dfrac{3/2}{x + 1} + \dfrac{(x + 1)/2}{x^{2} + 3}$ .
Cover-up method
$A = \left. \dfrac{P(x)}{(\text{denominator with factor }(x-a)\text{ removed})} \right|_{x = a}$
$a$
root of the factor $(x - a)$ being targeted
When to use
Fastest way to find the constant over a DISTINCT linear factor. Substitute $x = a$ into the full fraction after 'covering up' the $(x - a)$ in the denominator. NOT in the formula booklet — technique.
Example
In $\dfrac{7x - 1}{(x - 1)(x + 3)}$ , cover $(x - 1)$ and set $x = 1$ : $A = \dfrac{7(1) - 1}{1 + 3} = \dfrac{6}{4} = \dfrac{3}{2}$ .
Improper-fraction process
$\text{If } \deg P \geq \deg D: \;\; \dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)}, \;\; \text{then partial-fraction } \dfrac{R(x)}{D(x)}$
$Q(x)$
quotient (polynomial)
$R(x)$
remainder, $\deg R < \deg D$
When to use
Whenever the numerator degree $\geq$ denominator degree. Long-divide FIRST; then apply the standard partial-fraction templates to the proper-fraction remainder. NOT in the formula booklet.
Example
$\dfrac{x^{3} + 4x^{2} + 5x - 6}{(x - 1)(x + 2)} = (x + 3) + \dfrac{4x}{(x - 1)(x + 2)}$ .

Key Definitions and Keywords — Partial fractions

Definitions to memorise and the exact keywords mark schemes credit for partial fractions answers — 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
$\dfrac{5}{(x - 1)(x + 4)} = \dfrac{1}{x - 1} - \dfrac{1}{x + 4}$ .
Proper / improper rational function
Examiner keyword
A proper rational function has numerator degree strictly less than denominator degree. An improper one has $\deg$ num $\geq \deg$ denom. Partial-fraction templates apply only to proper fractions — improper ones must be polynomial-divided first.
Irreducible quadratic
Examiner keyword
A quadratic $ax^{2} + bx + c$ that cannot be factorised over the real numbers — equivalently, has discriminant $b^{2} - 4ac < 0$ .
Example
$x^{2} + 3$ is irreducible (no real roots). $x^{2} - 4 = (x - 2)(x + 2)$ 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 — Partial fractions

The traps other students keep falling into on partial fractions questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Writing only $\dfrac{B}{(x + 2)^{2}}$ for a repeated linear factor (forgetting the $\dfrac{A}{x + 2}$ term)
WMA14/01 examiner reports — flagged annually
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: $\dfrac{A}{x + 2} + \dfrac{B}{(x + 2)^{2}}$ . For $(x + 2)^{3}$ , add a third term $\dfrac{C}{(x + 2)^{3}}$ .
✕Using $\dfrac{B}{x^{2} + 3}$ instead of $\dfrac{Bx + C}{x^{2} + 3}$
WMA14/01 examiner reports — flagged annually
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.
✕Decomposing an improper fraction without long-dividing first
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.
✕Sign slip when applying cover-up at $x = -2$
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 $\dfrac{P(x)}{(x - 1)(x + 2)}$ at $x = -2$ : $(x - 1)$ becomes $-3$ , and the constant is $P(-2)$ . Track signs carefully.
✕Not verifying the decomposition by reassembling
Why it happens
Students stop after writing the answer.
How to avoid it
Combine your partial-fraction answer back over the common denominator — 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 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Partial fractions

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Distinct linear factors — cover-up method (4 marks, P4)

2Three distinct linear factors (5 marks, P4)

3Repeated linear factor (6 marks, P4)

4Irreducible quadratic factor (6 marks, P4)

5Improper fraction — divide first (6 marks, P4)

Distinct linear factors form

Repeated linear factor form

Irreducible quadratic factor form

Cover-up method

Improper-fraction process

Partial fractions

Proper / improper rational function

Irreducible quadratic

Cover-up method (Heaviside)

✕Writing only B(x+2)2\dfrac{B}{(x + 2)^{2}}(x+2)2B​ for a repeated linear factor (forgetting the Ax+2\dfrac{A}{x + 2}x+2A​ term)

✕Using Bx2+3\dfrac{B}{x^{2} + 3}x2+3B​ instead of Bx+Cx2+3\dfrac{Bx + C}{x^{2} + 3}x2+3Bx+C​

✕Decomposing an improper fraction without long-dividing first

✕Sign slip when applying cover-up at x=−2x = -2x=−2

✕Not verifying the decomposition by reassembling

Practice questions

Past paper style quiz

Video lesson

Partial fractions — frequently asked questions

✕Writing only $\dfrac{B}{(x + 2)^{2}}$ for a repeated linear factor (forgetting the $\dfrac{A}{x + 2}$ term)

✕Using $\dfrac{B}{x^{2} + 3}$ instead of $\dfrac{Bx + C}{x^{2} + 3}$

✕Sign slip when applying cover-up at $x = -2$