What is only one term for a repeated factor?

A factor (x - a)^2 needs both Ax - a and B(x - a)^2; one term alone cannot reproduce the original fraction.

What is a constant over a quadratic?

Over an irreducible quadratic the numerator is linear, Bx + C, not a single constant.

Express 7(x - 3)(x + 4) in partial fractions. [3 marks]

Write x(x - 2)^2 in partial fractions. [3 marks]

SQA AH style: distinct linear-style practice: Express 5x - 1(x - 1)(x + 2) in partial fractions.

Write 5x - 1(x - 1)(x + 2) = Ax - 1 + Bx + 2, so 5x - 1 = A(x + 2) + B(x - 1) (1 mark for the correct form). Cover-up at x = 1: 5(1) - 1 = A(3), so A = 43. Cover-up at x = -2: 5(-2) - 1 = B(-3), so -11 = -3B and B = 113 (1 mark). Answer: 43(x - 1) + 113(x + 2) (1 mark). Markers reward the correct form, the two constants, and the final expression.

SQA AH style: repeated factor-style practice: Express 3x + 4(x + 1)^2(x - 2) in partial fractions.

Form: Ax + 1 + B(x + 1)^2 + Cx - 2, so 3x + 4 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)^2 (1 mark). At x = -1: 1 = B(-3), so B = -13. At x = 2: 10 = C(9), so C = 109 (1 mark). Compare x^2 coefficients: 0 = A + C, so A = -109 (1 mark). Answer: -109(x + 1) - 13(x + 1)^2 + 109(x - 2) (1 mark). Markers reward the form with both repeated terms, the cover-up constants, and the coefficient comparison for A.

ScotlandMathsSyllabus dot point

How do you split a rational function into simpler fractions so it can be integrated or expanded?

Express a proper rational function as a sum of partial fractions where the denominator factorises into distinct linear factors, repeated linear factors, or an irreducible quadratic factor, and reduce an improper rational function first by algebraic division.

A focused answer to the SQA Advanced Higher Mathematics partial fractions content, covering proper and improper rational functions, denominators with distinct linear factors, repeated linear factors and an irreducible quadratic factor, and the cover-up and equating-coefficients methods used to find the constants.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Proper versus improper
The three denominator types
Finding the constants
Why this matters later
Try this

What this dot point is asking

The SQA wants you to take a rational function and rewrite it as a sum of simpler fractions. This is the gateway skill for integrating awkward fractions and for binomial expansions, so the algebra here pays off across the whole course.

Proper versus improper

A rational function is proper when the degree of the numerator is strictly less than the degree of the denominator. Only a proper fraction can be split into partial fractions directly. If the numerator degree is equal to or greater than the denominator degree, the fraction is improper and you must divide first, producing a polynomial part plus a proper remainder.

Reduce

\dfrac{x^2 + 3x + 1}{x^2 - 1}

before splitting

The degree of the numerator equals the degree of the denominator, making this an improper fraction. We must first reduce it by algebraic division to isolate a proper remainder, then split that remainder into partial fractions.

Step 1: Perform algebraic division to separate the whole part

Since both numerator and denominator have degree $2$ , the quotient is a constant. Divide $x^2 + 3x + 1$ by $x^2 - 1$ : the leading terms cancel with quotient $1$ , leaving a remainder $3x + 2$ .

\dfrac{x^2 + 3x + 1}{x^2 - 1} = 1 + \dfrac{3x + 2}{x^2 - 1}

because $x^2 + 3x + 1 = 1\cdot(x^2 - 1) + (3x + 2)$ .

Step 2: Factorise the denominator and set up the partial fractions

The remainder fraction now has $\deg(\text{numerator}) < \deg(\text{denominator})$ , so it is proper and can be split. Factorise the denominator first.

x^2 - 1 = (x - 1)(x + 1), \quad \text{so write} \quad \dfrac{3x + 2}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}

Step 3: Find $A$ and $B$ by the cover-up rule

Clear the denominators to get $3x + 2 = A(x + 1) + B(x - 1)$ . Substitute the roots of each linear factor to isolate each constant.

At $x = 1$ : $3(1) + 2 = A(2)$ , so $5 = 2A$ and $A = \dfrac{5}{2}$ .

At $x = -1$ : $3(-1) + 2 = B(-2)$ , so $-1 = -2B$ and $B = \dfrac{1}{2}$ .

Step 4: State the full decomposition

Combine the whole-number part with the partial fractions.

Final answer: $\dfrac{x^2 + 3x + 1}{x^2 - 1} = 1 + \dfrac{5}{2(x - 1)} + \dfrac{1}{2(x + 1)}$ .

The three denominator types

The shape of the decomposition is fixed by the factorisation of the denominator. Choose the form first, then find the constants.

The key rule for a repeated factor $(x - a)^2$ is that you need two terms, one over $(x - a)$ and one over $(x - a)^2$ . For an irreducible quadratic the numerator is a linear expression $Bx + C$ , not just a constant, because the numerator over any factor must have one degree less than that factor.

Finding the constants

Two methods combine well. The cover-up rule substitutes the root of a linear factor, which kills every other term and isolates one constant. Equating coefficients (or substituting convenient values) then finds any constants the cover-up cannot reach, such as the numerator over a quadratic or the non-squared term of a repeated factor.

Decompose

\dfrac{2x^2 + 1}{(x + 1)(x^2 + 1)}

The denominator has one linear factor $(x + 1)$ and one irreducible quadratic factor $(x^2 + 1)$ . The irreducible quadratic cannot be factorised over the reals, so it needs a linear numerator $Bx + C$ , not just a constant.

Step 1: Write the correct partial-fraction form

Assign a constant numerator to the linear factor and a linear numerator to the irreducible quadratic.

\dfrac{2x^2 + 1}{(x + 1)(x^2 + 1)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 1}

Multiplying through by $(x + 1)(x^2 + 1)$ gives the identity:

2x^2 + 1 = A(x^2 + 1) + (Bx + C)(x + 1)

Step 2: Use the cover-up rule to find $A$

Substitute $x = -1$ (the root of the linear factor). This kills the term containing $B$ and $C$ .

2(-1)^2 + 1 = A((-1)^2 + 1) \implies 3 = 2A \implies A = \dfrac{3}{2}

Step 3: Equate coefficients to find $B$ and $C$

The cover-up rule cannot isolate $B$ or $C$ directly, so expand the right side and compare coefficients of $x^2$ and of the constant term.

Coefficient of $x^2$ : $2 = A + B$ , so $B = 2 - \dfrac{3}{2} = \dfrac{1}{2}$ .

Constant term: $1 = A + C$ , so $C = 1 - \dfrac{3}{2} = -\dfrac{1}{2}$ .

Step 4: Write the final decomposition

Substitute $A$ , $B$ , $C$ back and simplify.

Final answer: $\dfrac{2x^2 + 1}{(x + 1)(x^2 + 1)} = \dfrac{3}{2(x + 1)} + \dfrac{x - 1}{2(x^2 + 1)}$ .

Why this matters later

Partial fractions are rarely the end of a question. They are the step that turns an unintegrable-looking fraction into a sum of standard integrals: $\dfrac{A}{x - a}$ integrates to $A\ln|x - a|$ , $\dfrac{B}{(x - a)^2}$ integrates to $-\dfrac{B}{x - a}$ , and a quadratic term splits into a logarithm and an inverse-tangent piece. They also let you expand a rational function as a binomial series term by term. Whenever you see a fraction you cannot integrate or expand directly, splitting it first is usually the intended route.

Try this

Q1. Express $\dfrac{7}{(x - 3)(x + 4)}$ in partial fractions. [3 marks]

Cue. Cover-up gives $A = 1$ and $B = -1$ , so $\dfrac{1}{x - 3} - \dfrac{1}{x + 4}$ .

Q2. Write $\dfrac{x}{(x - 2)^2}$ in partial fractions. [3 marks]

Cue. Form $\dfrac{A}{x - 2} + \dfrac{B}{(x - 2)^2}$ ; then $x = A(x - 2) + B$ gives $A = 1$ , $B = 2$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: distinct linear3 marksExpress

\dfrac{5x - 1}{(x - 1)(x + 2)}

in partial fractions.

Show worked answer →

Write $\dfrac{5x - 1}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ , so $5x - 1 = A(x + 2) + B(x - 1)$ (1 mark for the correct form).

Cover-up at $x = 1$ : $5(1) - 1 = A(3)$ , so $A = \dfrac{4}{3}$ . Cover-up at $x = -2$ : $5(-2) - 1 = B(-3)$ , so $-11 = -3B$ and $B = \dfrac{11}{3}$ (1 mark).

Answer: $\dfrac{4}{3(x - 1)} + \dfrac{11}{3(x + 2)}$ (1 mark). Markers reward the correct form, the two constants, and the final expression.

AH style: repeated factor4 marksExpress

\dfrac{3x + 4}{(x + 1)^2(x - 2)}

in partial fractions.

Show worked answer →

Form: $\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{x - 2}$ , so $3x + 4 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)^2$ (1 mark).

At $x = -1$ : $1 = B(-3)$ , so $B = -\dfrac{1}{3}$ . At $x = 2$ : $10 = C(9)$ , so $C = \dfrac{10}{9}$ (1 mark).

Compare $x^2$ coefficients: $0 = A + C$ , so $A = -\dfrac{10}{9}$ (1 mark).

Answer: $-\dfrac{10}{9(x + 1)} - \dfrac{1}{3(x + 1)^2} + \dfrac{10}{9(x - 2)}$ (1 mark). Markers reward the form with both repeated terms, the cover-up constants, and the coefficient comparison for $A$ .

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)