Polynomials, algebraic fractions and partial fractions - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on Polynomials, algebraic fractions and partial fractions for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

19 questions
3–10 marks each
Instant marking
No sign-up needed

19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Show that
\[\dfrac{1}{x - 1} - \dfrac{2}{x} + \dfrac{1}{x + 1} \equiv \dfrac{2}{x^3 - x}, \quad x ≠ 0, \ x ≠ \pm 1\]

Model answer

\(\dfrac{1}{x - 1} - \dfrac{2}{x} + \dfrac{1}{x + 1} = \dfrac{x(x + 1) - 2(x - 1)(x + 1) + x(x - 1)}{x(x - 1)(x + 1)}\)
Numerator: \(x^2 + x - 2x^2 + 2 + x^2 - x = 2\)
Denominator: \(x(x - 1)(x + 1) = x(x^2 - 1) = x^3 - x\)
So \(\dfrac{1}{x - 1} - \dfrac{2}{x} + \dfrac{1}{x + 1} \equiv \dfrac{2}{x^3 - x}\), as required.

Explanation

Use the common denominator \(x(x - 1)(x + 1)\), multiplying each numerator by the factors missing from its own denominator. The \(x^2\) and \(x\) terms cancel, leaving 2, and \(x(x - 1)(x + 1) = x^3 - x\). The common errors are sign slips when expanding \(-2(x - 1)(x + 1)\) and not showing that the denominator equals \(x^3 - x\).

Mark scheme

1 mark
M1 Combines the three fractions over a common denominator \(x(x - 1)(x + 1)\) (or \(x^3 - x\)), with each numerator multiplied by the missing factors. An answer with no working scores 0 marks
An answer with no working scores 0 marks: this is a 'show that' question. Alternative method: writes \(\dfrac{2}{x(x - 1)(x + 1)}\) in partial fractions as \(\dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x + 1}\) (M1), finds \(A = -2\), \(B = 1\), \(C = 1\) (A1), and concludes that this is the left-hand side (A1*).
1 mark
A1 Correct numerator, \(x(x + 1) - 2(x - 1)(x + 1) + x(x - 1)\) oe, correctly expanded: \(x^2 + x - 2x^2 + 2 + x^2 - x\)
1 mark
A1* Simplifies the numerator to 2 and shows that the denominator \(x(x - 1)(x + 1) = x(x^2 - 1) = x^3 - x\), reaching \(\dfrac{2}{x^3 - x}\) with no errors
cso. The denominator must be shown to equal \(x^3 - x\). A sign error in \(-2(x^2 - 1)\), e.g. \(-2x^2 - 2\), scores A0 A0*.

Question 2

4 marks

Given that
\[\dfrac{6x^2 - x + 2}{(2x - 1)(x + 1)} \equiv A + \dfrac{B}{2x - 1} + \dfrac{C}{x + 1}\]
find the values of the constants \(A\), \(B\) and \(C\).

Model answer

\(6x^2 - x + 2 \equiv A(2x - 1)(x + 1) + B(x + 1) + C(2x - 1)\)
Coefficients of \(x^2\): \(6 = 2A\), so \(A = 3\)
\(x = \dfrac{1}{2}\): \(\dfrac{3}{2} - \dfrac{1}{2} + 2 = \dfrac{3}{2}B\), so \(3 = \dfrac{3}{2}B\) and \(B = 2\)
\(x = -1\): \(6 + 1 + 2 = -3C\), so \(C = -3\)
\(A = 3\), \(B = 2\), \(C = -3\)

Explanation

The numerator and denominator both have degree 2, so the fraction is improper and needs the constant \(A\) as well as the two partial fractions. Multiplying through gives an identity in which the \(x^2\) coefficients give \(A\), and substituting \(x = \dfrac{1}{2}\) and \(x = -1\) gives \(B\) and \(C\). The most common error is leaving out the \(A(2x - 1)(x + 1)\) term in the identity.

Mark scheme

1 mark
B1 \(A = 3\); correct values \(A = 3\), \(B = 2\), \(C = -3\) with no working score all 4 marks
An answer of \(A = 3\), \(B = 2\) and \(C = -3\) with no working scores 4 marks. \(A\) comes from comparing the \(x^2\) terms (\(6 = 2A\)) or from algebraic division: \(6x^2 - x + 2 = 3(2x^2 + x - 1) - 4x + 5\).
1 mark
M1 Forms \(6x^2 - x + 2 \equiv A(2x - 1)(x + 1) + B(x + 1) + C(2x - 1)\) (or uses their remainder: \(-4x + 5 \equiv B(x + 1) + C(2x - 1)\)) and substitutes a value of \(x\) or compares coefficients to find \(B\) or \(C\)
An identity without the \(A(2x - 1)(x + 1)\) term (treating the fraction as if it were proper) scores M0.
1 mark
A1 \(B = 2\)
1 mark
A1 \(C = -3\)
Depends on the M mark.

Question 3

5 marks

In this question you must show all stages of your working.

Solve the equation
\[\dfrac{x}{x - 2} + \dfrac{4}{x + 1} = \dfrac{6}{x^2 - x - 2}\]

Model answer

\(x^2 - x - 2 = (x - 2)(x + 1)\)
Multiply every term by \((x - 2)(x + 1)\):
\(x(x + 1) + 4(x - 2) = 6\)
\(x^2 + x + 4x - 8 - 6 = 0\)
\(x^2 + 5x - 14 = 0\)
\((x + 7)(x - 2) = 0\), so \(x = -7\) or \(x = 2\)
\(x = 2\) makes the denominators \(x - 2\) and \(x^2 - x - 2\) zero, so it is not a solution.
The only solution is \(x = -7\).

Explanation

Factorising \(x^2 - x - 2\) as \((x - 2)(x + 1)\) shows the common denominator, and multiplying through clears the fractions to give a quadratic. One root of the quadratic, \(x = 2\), makes the original denominators zero, so it must be rejected, leaving \(x = -7\). Common errors are not multiplying every term by the common denominator and failing to check the roots against the original equation.

Mark scheme

1 mark
M1 Factorises \(x^2 - x - 2 = (x - 2)(x + 1)\) and multiplies every term by \((x - 2)(x + 1)\) (or writes the left-hand side over this common denominator). An answer with no working scores 0 marks
An answer of \(x = -7\) with no working scores 0 marks: the question demands all stages of working.
1 mark
A1 \(x(x + 1) + 4(x - 2) = 6\) oe
1 mark
A1 \(x^2 + 5x - 14 = 0\)
1 mark
M1 (dM1) Solves their three-term quadratic
Depends on the first M mark.
1 mark
A1 \(x = -7\) only, with \(x = 2\) rejected because it makes a denominator zero
Giving \(x = 2\) as a solution, or giving both \(-7\) and 2 without rejecting 2, scores A0. A reason is needed for rejecting \(x = 2\), e.g. 'the equation is undefined when \(x = 2\)'.

Question 4

7 marks

\[\mathrm{f}(x) = \dfrac{5 + x}{(1 - x)(1 + 2x)}, \quad x ≠ 1, \ x ≠ -\dfrac{1}{2}\]
Express \(\mathrm{f}(x)\) in partial fractions.

Hence find, in terms of \(r\), the coefficient of \(x^r\) in the series expansion of \(\mathrm{f}(x)\) in ascending powers of \(x\), where \(r\) is a non-negative integer.

Find the value of \(r\) for which this coefficient is 194, and state the set of values of \(x\) for which the expansion is valid.

[You may use \((1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{1 \times 2}x^2 + \ldots + \dfrac{n(n - 1)\ldots(n - r + 1)}{1 \times 2 \times \ldots \times r}x^r + \ldots\), valid for \(|x| < 1\), \(n \in \mathbb{R}\).]

Model answer

\(5 + x \equiv A(1 + 2x) + B(1 - x)\)
\(x = 1\): \(6 = 3A\), so \(A = 2\)
\(x = -\dfrac{1}{2}\): \(\dfrac{9}{2} = \dfrac{3}{2}B\), so \(B = 3\)
\(\mathrm{f}(x) = \dfrac{2}{1 - x} + \dfrac{3}{1 + 2x} = 2(1 - x)^{-1} + 3(1 + 2x)^{-1}\)
In \((1 + x)^{-1}\) the coefficient of \(x^r\) is \(\dfrac{(-1)(-2)\ldots(-r)}{1 \times 2 \times \ldots \times r} = (-1)^r\).
Replacing \(x\) by \(-x\): the coefficient of \(x^r\) in \((1 - x)^{-1}\) is \((-1)^r(-1)^r = 1\), so \(2(1 - x)^{-1} = 2 + 2x + 2x^2 + \ldots + 2x^r + \ldots\)
Replacing \(x\) by \(2x\): the coefficient of \(x^r\) in \((1 + 2x)^{-1}\) is \((-1)^r 2^r = (-2)^r\), so \(3(1 + 2x)^{-1} = 3 - 6x + 12x^2 - \ldots + 3(-2)^r x^r + \ldots\)
So the coefficient of \(x^r\) in the expansion of \(\mathrm{f}(x)\) is \(2 + 3(-2)^r\).
\(2 + 3(-2)^r = 194\), so \((-2)^r = 64\). This needs \(r\) even with \(2^r = 64\), so \(r = 6\).
\((1 - x)^{-1}\) is valid for \(|x| < 1\) and \((1 + 2x)^{-1}\) for \(|2x| < 1\), so the expansion of \(\mathrm{f}(x)\) is valid for \(|x| < \dfrac{1}{2}\).

Explanation

Writing \(\mathrm{f}(x)\) as \(2(1 - x)^{-1} + 3(1 + 2x)^{-1}\) lets you find the general term of each expansion separately: the terms are \(2x^r\) and \(3(-2x)^r\), so the coefficient of \(x^r\) is \(2 + 3(-2)^r\), and setting this equal to 194 gives \((-2)^r = 64\), so \(r = 6\). The expansion is valid only where both series converge, which is the narrower range \(|x| < \dfrac{1}{2}\). Common errors are losing the sign in \((-2)^r\), writing out a few terms without a general term, and giving \(|x| < 1\) as the range.

Mark scheme

1 mark
M1 Uses \(\dfrac{A}{1 - x} + \dfrac{B}{1 + 2x}\) with \(5 + x \equiv A(1 + 2x) + B(1 - x)\) and finds one constant by substituting a value of \(x\) or comparing coefficients; or correct final answers with no working (5 marks, see comments)
An answer of \(\dfrac{2}{1 - x} + \dfrac{3}{1 + 2x}\), coefficient \(2 + 3(-2)^r\), \(r = 6\) and \(|x| < \dfrac{1}{2}\) with no working scores 5 marks: M1 A1 for the partial fractions, the final A1 for the coefficient, and both B1 marks; the M1 and A1 for the general terms need to be seen. An identity such as \(5 + x \equiv A(1 - x) + B(1 + 2x)\) scores M0.
1 mark
A1 \(\mathrm{f}(x) = \dfrac{2}{1 - x} + \dfrac{3}{1 + 2x}\)
The partial fractions must be written out: the values \(A = 2\) and \(B = 3\) alone score A0.
1 mark
M1 Writes one of their partial fractions as a power, e.g. \(2(1 - x)^{-1}\) or \(3(1 + 2x)^{-1}\), and finds the general term of its expansion, e.g. the coefficient of \(x^r\) in \((1 + 2x)^{-1}\) is \(\dfrac{(-1)(-2)\ldots(-r)}{1 \times 2 \times \ldots \times r} \times 2^r = (-2)^r\)
Follow through their \(A\) and \(B\). Alternative: recognises a geometric series, e.g. \((1 + 2x)^{-1} = 1 - 2x + (2x)^2 - \ldots\) with common ratio \(-2x\), so its general term is \((-2x)^r\). Writing only the first few terms with no general term scores M0.
1 mark
A1 Both general terms correct: \(2x^r\) from \(2(1 - x)^{-1}\) and \(3(-2)^r x^r\) from \(3(1 + 2x)^{-1}\)
A common error is \(3 \times 2^r x^r\) for the second general term, from replacing \(x\) by \(2x\) and losing the sign: A0.
1 mark
A1 The coefficient of \(x^r\) is \(2 + 3(-2)^r\)
Accept equivalent forms, e.g. \(2 + 3 \times (-1)^r 2^r\).
1 mark
B1 \(r = 6\)
From \(2 + 3(-2)^r = 194\), \((-2)^r = 64\), so \(r\) is even and \(2^r = 64\). Follow through is not allowed: the answer must be 6.
1 mark
B1 \(|x| < \dfrac{1}{2}\)
Accept \(-\dfrac{1}{2} < x < \dfrac{1}{2}\), \(\{x : -\dfrac{1}{2} < x < \dfrac{1}{2}\}\) and typed forms such as |x| < 1/2. \(|x| < 1\) or \(|x| \leqslant \dfrac{1}{2}\) scores B0: both expansions must be valid, and \((1 + 2x)^{-1}\) needs \(|2x| < 1\).

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.