Partial Fractions (AA HL)
Splitting a single algebraic fraction into a sum of simpler ones is a skill that shows up constantly once you reach HL calculus - integrating a rational function, for instance, is usually impossible until it's been broken into partial fractions first. This page focuses on the cover-up method and coefficient comparison for denominators with up to two distinct linear factors, with worked examples and the mistakes that lose the most marks. It's part of the broader Polynomials & Partial Fractions topic.
24 questions on this sub-topic.
The partial fraction identity
Covered under IB syllabus reference AHL1.11: partial fractions with a maximum of two distinct linear terms in the denominator, and the degree of the numerator less than the degree of the denominator.
Distinct linear factors
\(\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\)
Not in the formula booklet - it's a method you're expected to know, not a formula to look up.
Cover-up method
Multiply the identity by one denominator factor and substitute the value of \(x\) that makes it zero.
Finds each constant directly - much faster than expanding and comparing coefficients throughout.
Need the full syllabus wording and formula-booklet reference table? See Polynomials & Partial Fractions. For GDC-based checks of a partial fraction answer, see the parent topic's GDC guidance.
Worked examples
In the identity \(\dfrac{7}{(x-3)(x+4)}=\dfrac{A}{x-3}+\dfrac{B}{x+4}\), find \(A\).
Worked solution
Multiply by \((x-3)\) and set \(x = 3\) (cover-up): M1
\(A = \dfrac{7}{3+4} = \dfrac{7}{7} = 1.\) A1
Express \(\dfrac{5x - 4}{(x - 2)(x + 1)}\) in partial fractions.
(a) State \(A.\)
(b) State \(B.\)
Worked solution
\(\dfrac{5x-4}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}.\) M1 So \(5x - 4 = A(x+1) + B(x-2).\) A1
\(x = 2\): \(6 = 3A \Rightarrow A = 2.\) M1 A1 \(x = -1\): \(-9 = -3B \Rightarrow B = 3.\) A1
Express \(\dfrac{2x^2 - 3x + 4}{x(x-1)^2}\) in partial fractions.
(a) State \(A.\)
(b) State \(B.\)
(c) State \(C.\)
Worked solution
(a) \(\dfrac{2x^2-3x+4}{x(x-1)^2} = \dfrac{A}{x} + \dfrac{B}{x-1} + \dfrac{C}{(x-1)^2}.\) \(2x^2-3x+4 = A(x-1)^2 + Bx(x-1) + Cx.\) M1
\(2x^2-3x+4 = A(x-1)^2 + Bx(x-1) + Cx.\) A1
\(x=0\): \(4 = A.\) M1
\(A=4.\) A1
(b) Compare coefficients of \(x^2\): \(2 = A + B.\) M1 \(B = -2.\) So \(\dfrac{4}{x} - \dfrac{2}{x-1} + \dfrac{3}{(x-1)^2}.\) A1
(c) \(x=1\): \(3 = C.\) A1
Common mistakes
- Not checking the partial fraction identity holds for all \(x\). After finding \(A\) and \(B\), it's worth checking with one more value of \(x\) (or by comparing a coefficient) that the identity actually balances.
- Using the wrong form for a repeated factor. A denominator of \((x-a)^2\) needs two separate terms, \(\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}\), not a single \(\dfrac{A}{(x-a)^2}\) - missing the first term makes the decomposition impossible to balance.
- Cover-up only finds some of the constants. The cover-up method gives you a constant instantly whenever its factor is set to zero, but for a repeated factor or an irreducible quadratic factor, at least one constant still needs a coefficient comparison - don't assume cover-up alone will finish the question.
Ready to practise properly?
25 partial-fractions questions, marked instantly like the real exam.
Quick answers
What is the cover-up method for partial fractions?
For a distinct linear factor \((x-a)\) in the denominator, multiply through by that factor and substitute \(x=a\). This "covers up" the \((x-a)\) term and leaves you with the corresponding numerator constant directly.
How do partial fractions work with a repeated linear factor?
A repeated factor \((x-a)^2\) needs two terms in the decomposition: \(\dfrac{A}{x-a}\) and \(\dfrac{B}{(x-a)^2}\). The cover-up method finds \(B\) directly at \(x=a\), but \(A\) usually needs a coefficient comparison.