AS & A Level · Practice questions · Partial fractions

Split it, then use it.

Paper 3 only. Six original Cambridge-style questions on partial fractions, covering all three syllabus denominator types and finishing with a split-then-differentiate problem.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Form, identity, substitute, compare.

Factorise the denominator fully, and compare degrees: equal degrees need a constant term \(A\) as well.
Form: one fraction per linear factor; \(\dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^2}\) for a repeated factor; \(\dfrac{Bx + C}{cx^2 + d}\) for a quadratic factor.
Identity: multiply through by the whole denominator.
Substitute the root of each linear factor to find its constant directly.
Compare coefficients (usually \(x^2\) and constants) for the rest, then check one more coefficient.

Recall

[3 marks]

Write down the form of the partial fractions for each of the following. Do not find the constants.

(a) \(\dfrac{x + 1}{(x - 2)(3x + 1)}\) (b) \(\dfrac{2x^2 + 1}{(x + 4)^2(x - 5)}\) (c) \(\dfrac{x}{(2x - 1)(x^2 + 5)}\)

(a) \(\dfrac{A}{x - 2} + \dfrac{B}{3x + 1}\) ✓
(b) \(\dfrac{A}{x - 5} + \dfrac{B}{x + 4} + \dfrac{C}{(x + 4)^2}\) ✓
(c) \(\dfrac{A}{2x - 1} + \dfrac{Bx + C}{x^2 + 5}\) ✓

Calculation

[4 marks]

Express \(\dfrac{x^2 + 4x - 2}{(x - 2)(x + 3)}\) in partial fractions.

Numerator and denominator both have degree 2, so the form is \(A + \dfrac{B}{x - 2} + \dfrac{C}{x + 3}\). ✓
\(x^2 + 4x - 2 \equiv A(x - 2)(x + 3) + B(x + 3) + C(x - 2)\). Compare \(x^2\): \(A = 1\). ✓
\(x = 2\): \(10 = 5B\), so \(B = 2\). \(x = -3\): \(-5 = -5C\), so \(C = 1\). ✓
\(1 + \dfrac{2}{x - 2} + \dfrac{1}{x + 3}\) ✓

Calculation

[5 marks]

Express \(\dfrac{9x^2 + 5x - 10}{(x - 1)(x + 3)(2x - 1)}\) in partial fractions.

\(9x^2 + 5x - 10 \equiv A(x + 3)(2x - 1) + B(x - 1)(2x - 1) + C(x - 1)(x + 3)\). ✓
\(x = 1\): \(4 = A(4)(1)\), so \(A = 1\). ✓
\(x = -3\): \(81 - 15 - 10 = 56 = B(-4)(-7) = 28B\), so \(B = 2\). ✓
\(x = \dfrac{1}{2}\): \(\dfrac{9}{4} + \dfrac{5}{2} - 10 = -\dfrac{21}{4} = C\left(-\dfrac{1}{2}\right)\left(\dfrac{7}{2}\right) = -\dfrac{7}{4}C\), so \(C = 3\). ✓
\(\dfrac{1}{x - 1} + \dfrac{2}{x + 3} + \dfrac{3}{2x - 1}\) (check \(x^2\): \(2A + 2B + C = 2 + 4 + 3 = 9\)). ✓

Calculation

[5 marks]

Express \(\dfrac{4 + 6x - x^2}{(2x - 1)(x + 1)^2}\) in partial fractions.

\(4 + 6x - x^2 \equiv A(x + 1)^2 + B(2x - 1)(x + 1) + C(2x - 1)\). ✓
\(x = \dfrac{1}{2}\): \(4 + 3 - \dfrac{1}{4} = \dfrac{27}{4} = \dfrac{9}{4}A\), so \(A = 3\). ✓
\(x = -1\): \(4 - 6 - 1 = -3 = -3C\), so \(C = 1\). ✓
\(x^2\): \(-1 = A + 2B\), so \(B = -2\). ✓
\(\dfrac{3}{2x - 1} - \dfrac{2}{x + 1} + \dfrac{1}{(x + 1)^2}\) ✓

Calculation

[5 marks]

Express \(\dfrac{5x^2 + 11x}{(3x - 1)(x^2 + 2)}\) in partial fractions.

\(5x^2 + 11x \equiv A(x^2 + 2) + (Bx + C)(3x - 1)\). ✓
\(x = \dfrac{1}{3}\): \(\dfrac{5}{9} + \dfrac{11}{3} = \dfrac{38}{9} = A\left(\dfrac{19}{9}\right)\), so \(A = 2\). ✓
\(x^2\): \(5 = A + 3B\), so \(B = 1\). ✓
Constants: \(0 = 2A - C\), so \(C = 4\). ✓
\(\dfrac{2}{3x - 1} + \dfrac{x + 4}{x^2 + 2}\) (check \(x\): \(3C - B = 12 - 1 = 11\)). ✓

Analysis

[7 marks]

Let \(f(x) = \dfrac{x^2 + 8x + 11}{(x + 1)^2(x + 3)}\).

(a) Express \(f(x)\) in partial fractions.

(b) Hence find \(f'(x)\), and show that \(f'(0) = -\dfrac{53}{9}\).

(a) \(x^2 + 8x + 11 \equiv A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2\), with \(f(x) = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{x + 3}\). ✓
\(x = -1\): \(4 = 2B\), so \(B = 2\). \(x = -3\): \(-4 = 4C\), so \(C = -1\). ✓
\(x^2\): \(1 = A + C\), so \(A = 2\). So \(f(x) = \dfrac{2}{x + 1} + \dfrac{2}{(x + 1)^2} - \dfrac{1}{x + 3}\). ✓
(b) Write \(f(x) = 2(x + 1)^{-1} + 2(x + 1)^{-2} - (x + 3)^{-1}\) and differentiate each term with the chain rule. ✓
\(f'(x) = -\dfrac{2}{(x + 1)^2} - \dfrac{4}{(x + 1)^3} + \dfrac{1}{(x + 3)^2}\). ✓
\(f'(0) = -2 - 4 + \dfrac{1}{9} = -\dfrac{53}{9}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.