P2.10Partial fractions

Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Express \(\frac{-x - 5}{(x + 3)(x + 4)}\) in partial fractions.[3]

Show the answer and mark scheme

Answer: \(-\frac{2}{x + 3} + \frac{1}{x + 4}\)

M1 for the form \(\frac{A}{x + 3} + \frac{B}{x + 4}\) and a correct method to find a constant (substitution, cover-up or equating coefficients)
A1 for one correct constant
A1 for \(-\frac{2}{x + 3} + \frac{1}{x + 4}\)

Worked solution: \(-x - 5 \equiv A(x + 4) + B(x + 3)\).
Let \(x = -4\): \(-1 = -B\), so \(B = 1\).
Let \(x = -3\): \(-2 = A\), so \(A = -2\).
So the expression is \(-\frac{2}{x + 3} + \frac{1}{x + 4}\).

Question 2Medium4 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Express \(\frac{-10x^{2} - 3x + 27}{x(x + 3)(x - 3)}\) in partial fractions.[4]

Show the answer and mark scheme

Answer: \(-\frac{3}{x + 3} - \frac{3}{x} - \frac{4}{x - 3}\)

M1 for the form \(\frac{A}{x + 3} + \frac{B}{x} + \frac{C}{x - 3}\) and a correct method to find the constants
A1 for one correct constant
A1 for a second correct constant
A1 for \(-\frac{3}{x + 3} - \frac{3}{x} - \frac{4}{x - 3}\)

Worked solution: \(-10x^{2} - 3x + 27 \equiv Ax(x - 3) + B(x + 3)(x - 3) + Cx(x + 3)\).
Let \(x = -3\): \(-54 = (-3) \times (-6) \times A\), so \(A = -3\).
Let \(x = 0\): \(27 = 3 \times (-3) \times B\), so \(B = -3\).
Let \(x = 3\): \(-72 = 6 \times 3 \times C\), so \(C = -4\).
So the expression is \(-\frac{3}{x + 3} - \frac{3}{x} - \frac{4}{x - 3}\).

Question 3Hard5 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Express \(\frac{8x^{2} + 41x + 48}{4x^{3} - 27x - 27}\) in partial fractions.[5]

Show the answer and mark scheme

Answer: \(\frac{3}{x - 3} - \frac{2}{2x + 3} - \frac{1}{(2x + 3)^2}\)

B1 for factorising the denominator: \((x - 3)(2x + 3)^2\)
M1 for the form \(\frac{A}{x - 3} + \frac{B}{2x + 3} + \frac{C}{(2x + 3)^2}\)
M1 for a correct method to find the constants, e.g. substituting \(x = 3\) and \(x = -\frac{3}{2}\) and comparing coefficients of \(x^2\)
A1 for two correct constants
A1 for \(\frac{3}{x - 3} - \frac{2}{2x + 3} - \frac{1}{(2x + 3)^2}\)

Worked solution: \(4x^{3} - 27x - 27 = (x - 3)(2x + 3)^2\).
\(8x^{2} + 41x + 48 \equiv A(2x + 3)^2 + B(x - 3)(2x + 3) + C(x - 3)\).
Let \(x = 3\): \(243 = 81A\), so \(A = 3\).
Let \(x = -\frac{3}{2}\): \(\frac{9}{2} = -\frac{9}{2}C\), so \(C = -1\).
Comparing coefficients of \(x^2\): \(8 = 4A + 2B\), so \(B = -2\).
So the expression is \(\frac{3}{x - 3} - \frac{2}{2x + 3} - \frac{1}{(2x + 3)^2}\).

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