Polynomials, algebraic fractions and partial fractions - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Polynomials, algebraic fractions and partial fractions for Edexcel A-Level Maths (Paper 2). 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.

18 questions
3–10 marks each
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18 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

Express
\[\dfrac{x + 7}{x^2 - x - 6} - \dfrac{2}{x - 3}, \quad x ≠ 3, \ x ≠ -2\]
as a single fraction in its simplest form.

Model answer

\(x^2 - x - 6 = (x - 3)(x + 2)\)
\(\dfrac{x + 7}{(x - 3)(x + 2)} - \dfrac{2(x + 2)}{(x - 3)(x + 2)} = \dfrac{x + 7 - 2x - 4}{(x - 3)(x + 2)} = \dfrac{3 - x}{(x - 3)(x + 2)}\)
\(3 - x = -(x - 3)\), so the fraction is \(-\dfrac{1}{x + 2}\)

Explanation

Factorising \(x^2 - x - 6\) shows that \((x - 3)(x + 2)\) is the lowest common denominator, so the second fraction needs to be multiplied top and bottom by \((x + 2)\). The numerator simplifies to \(3 - x = -(x - 3)\), which cancels with the factor \((x - 3)\) in the denominator. Common errors are a sign slip with \(-2(x + 2)\) and not spotting that \(3 - x\) and \(x - 3\) cancel.

Mark scheme

1 mark
M1 Factorises \(x^2 - x - 6 = (x - 3)(x + 2)\) and writes both fractions over a common denominator, e.g. \(\dfrac{x + 7 - 2(x + 2)}{(x - 3)(x + 2)}\); or a correct final answer of \(-\dfrac{1}{x + 2}\) with no working
An answer of \(-\dfrac{1}{x + 2}\) with no working scores 3 marks. A common denominator of \((x^2 - x - 6)(x - 3)\) is also acceptable for this mark if it is used correctly.
1 mark
A1 \(\dfrac{3 - x}{(x - 3)(x + 2)}\) oe
A common error is \(x + 7 - 2x + 4\), from expanding \(-2(x + 2)\) as \(-2x + 4\): A0.
1 mark
A1 \(-\dfrac{1}{x + 2}\)
Depends on the M mark. Accept \(\dfrac{-1}{x + 2}\) and \(\dfrac{1}{-x - 2}\). \(\dfrac{3 - x}{(x - 3)(x + 2)}\) is not in its simplest form: A0 for this mark.

Question 2

5 marks

Express
\[\dfrac{3x^2 + 5x + 10}{(x^2 - 4)(x + 2)}, \quad x ≠ \pm 2\]
in partial fractions.

Model answer

\(x^2 - 4 = (x - 2)(x + 2)\), so the denominator is \((x - 2)(x + 2)^2\)
\(\dfrac{3x^2 + 5x + 10}{(x - 2)(x + 2)^2} \equiv \dfrac{A}{x - 2} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^2}\)
\(3x^2 + 5x + 10 \equiv A(x + 2)^2 + B(x - 2)(x + 2) + C(x - 2)\)
\(x = 2\): \(12 + 10 + 10 = 16A\), so \(A = 2\)
\(x = -2\): \(12 - 10 + 10 = -4C\), so \(C = -3\)
Coefficients of \(x^2\): \(3 = A + B\), so \(B = 1\)
\(\dfrac{3x^2 + 5x + 10}{(x^2 - 4)(x + 2)} = \dfrac{2}{x - 2} + \dfrac{1}{x + 2} - \dfrac{3}{(x + 2)^2}\)

Explanation

Factorising \(x^2 - 4\) shows that \((x + 2)\) occurs twice, so the denominator is \((x - 2)(x + 2)^2\) and needs partial fractions over \((x - 2)\), \((x + 2)\) and \((x + 2)^2\). Substituting \(x = 2\) and \(x = -2\) gives A and C, and comparing the \(x^2\) coefficients gives B. Common errors are missing the repeated factor and writing three fractions with linear denominators, or putting a constant over \(x^2 - 4\).

Mark scheme

1 mark
B1 Writes \(x^2 - 4 = (x - 2)(x + 2)\), so that the denominator is \((x - 2)(x + 2)^2\), and uses a correct form \(\dfrac{A}{x - 2} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^2}\); a correct final answer of \(\dfrac{2}{x - 2} + \dfrac{1}{x + 2} - \dfrac{3}{(x + 2)^2}\) with no working scores all 5 marks
An answer of \(\dfrac{2}{x - 2} + \dfrac{1}{x + 2} - \dfrac{3}{(x + 2)^2}\) with no working scores 5 marks. A common error is to treat the denominator as three different linear factors, writing \(\dfrac{A}{x - 2} + \dfrac{B}{x + 2} + \dfrac{C}{x + 2}\), or to use \(\dfrac{A}{x^2 - 4} + \dfrac{B}{x + 2}\): these forms score B0 and cannot lead to a correct answer.
1 mark
M1 Forms a correct identity for their form, e.g. \(3x^2 + 5x + 10 \equiv A(x + 2)^2 + B(x - 2)(x + 2) + C(x - 2)\), and substitutes a value of x or compares coefficients to find a constant
An identity in which C is multiplied by \((x - 2)(x + 2)\) scores M0.
1 mark
A1 One correct constant: \(A = 2\), \(B = 1\) or \(C = -3\)
1 mark
A1 A second correct constant
Depends on the M mark.
1 mark
A1 \(\dfrac{2}{x - 2} + \dfrac{1}{x + 2} - \dfrac{3}{(x + 2)^2}\)
Depends on the M mark. All three constants correct and the partial fractions written out; accept \(+ \dfrac{-3}{(x + 2)^2}\).

Question 3

6 marks

Find the values of the constants \(p\) and \(q\) such that
\[x^4 + 4x^3 - 4x^2 - 16x + 16 \equiv \left(x^2 + px + q\right)^2\]
Hence solve the equation
\[x^4 + 4x^3 - 4x^2 - 16x + 16 = 0\]
giving your answers in the form \(a \pm \sqrt{b}\), where \(a\) and \(b\) are integers, and explain why \(x^4 + 4x^3 - 4x^2 - 16x + 16 \geqslant 0\) for all real values of \(x\).

Model answer

\(\left(x^2 + px + q\right)^2 = x^4 + 2px^3 + (p^2 + 2q)x^2 + 2pqx + q^2\)
Coefficients of \(x^3\): \(2p = 4\), so \(p = 2\)
Coefficients of \(x^2\): \(p^2 + 2q = -4\), so \(4 + 2q = -4\) and \(q = -4\)
Check: \(2pq = 2(2)(-4) = -16\) and \(q^2 = 16\), as required.
\(x^4 + 4x^3 - 4x^2 - 16x + 16 = \left(x^2 + 2x - 4\right)^2 = 0 \iff x^2 + 2x - 4 = 0\)
\((x + 1)^2 = 5\), so \(x = -1 \pm \sqrt{5}\)
The expression is the square of the real number \(x^2 + 2x - 4\), so it is never negative: \(x^4 + 4x^3 - 4x^2 - 16x + 16 \geqslant 0\) for all real \(x\).

Explanation

Expanding the square and comparing coefficients gives \(p\) from the \(x^3\) terms and then \(q\) from the \(x^2\) terms, and the \(x\) and constant terms check the result. Once the quartic is written as \(\left(x^2 + 2x - 4\right)^2\), it is zero only when the quadratic inside is zero, giving \(x = -1 \pm \sqrt{5}\) (each a repeated root), and as a square it can never be negative. Common errors are taking \(q = 4\) from \(q^2 = 16\) without checking the other coefficients and solving the quartic on a calculator.

Mark scheme

1 mark
M1 Expands \(\left(x^2 + px + q\right)^2\) to \(x^4 + 2px^3 + (p^2 + 2q)x^2 + 2pqx + q^2\) (or enough of it) and compares coefficients; correct answers with no working score 5 marks (see comments)
An answer of \(p = 2\), \(q = -4\) and \(x = -1 \pm \sqrt{5}\) with no working scores 5 marks: the last mark needs the explanation.
1 mark
A1 \(p = 2\)
From \(2p = 4\).
1 mark
A1 \(q = -4\)
Depends on the M mark. From \(p^2 + 2q = -4\); it must be consistent with \(2pq = -16\) and \(q^2 = 16\). \(q = \pm 4\) or \(q = 4\) scores A0.
1 mark
M1 Uses the identity to solve \(x^2 + 2x - 4 = 0\) (their \(x^2 + px + q = 0\)) by the formula or completing the square
Not dependent on the earlier marks. Solving the quartic by calculator scores M0 here, as the question says 'hence'.
1 mark
A1 \(x = -1 + \sqrt{5}\) and \(x = -1 - \sqrt{5}\)
Accept \(x = -1 \pm \sqrt{5}\). Decimals alone (1.24, \(-3.24\)) score A0.
1 mark
B1 Explains that the expression equals \(\left(x^2 + 2x - 4\right)^2\), the square of a real number, so it is never negative (it is 0 only at \(x = -1 \pm \sqrt{5}\))
Accept 'it is a perfect square, so it is \(\geqslant 0\)'. A statement that squares are 'always positive' with no allowance for 0 is condoned here only if \(\geqslant 0\) is concluded.

Question 4

8 marks

\[\mathrm{f}(x) = \dfrac{9 - 14x}{(1 - 4x)(3 - x)}, \quad \lvert x \rvert < \dfrac{1}{4}\]
Express \(\mathrm{f}(x)\) in partial fractions.

Hence find the series expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), up to and including the term in \(x^2\), giving each coefficient as an integer or as a fraction in its simplest form.

For \(0 < x < \dfrac{1}{4}\), the first three terms of this expansion are used as an approximation to \(\mathrm{f}(x)\). State, with a reason, whether this approximation is an overestimate or an underestimate of \(\mathrm{f}(x)\).

[The binomial series: \((1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{1 \times 2}x^2 + \dots + \dfrac{n(n - 1)\dots(n - r + 1)}{1 \times 2 \times \dots \times r}x^r + \dots, \quad |x| < 1, \ n \in \mathbb{R}\)]

Model answer

\(9 - 14x \equiv A(3 - x) + B(1 - 4x)\)
\(x = \dfrac{1}{4}\): \(9 - \dfrac{7}{2} = \dfrac{11}{4}A\), so \(\dfrac{11}{2} = \dfrac{11}{4}A\) and \(A = 2\)
\(x = 3\): \(9 - 42 = -11B\), so \(B = 3\)
\(\mathrm{f}(x) = \dfrac{2}{1 - 4x} + \dfrac{3}{3 - x}\)
\(2(1 - 4x)^{-1} = 2\left(1 + (-1)(-4x) + \dfrac{(-1)(-2)}{2}(-4x)^2 + \dots\right) = 2 + 8x + 32x^2 + \dots\)
\(\dfrac{3}{3 - x} = \left(1 - \dfrac{x}{3}\right)^{-1} = 1 + \dfrac{x}{3} + \dfrac{x^2}{9} + \dots\)
\(\mathrm{f}(x) = 3 + \dfrac{25}{3}x + \dfrac{289}{9}x^2 + \dots\)
The next terms are \(128x^3\) and \(\dfrac{1}{27}x^3\); in general the terms are \(2(4x)^r\) and \(\left(\dfrac{x}{3}\right)^r\), so for \(0 < x < \dfrac{1}{4}\) every term left out is positive and their sum is positive.
So the approximation is an underestimate of \(\mathrm{f}(x)\).

Explanation

Partial fractions split \(\mathrm{f}(x)\) into \(2(1 - 4x)^{-1} + 3(3 - x)^{-1}\), and the second term must be rewritten as \(\left(1 - \dfrac{x}{3}\right)^{-1}\) before the binomial series is used; adding the two expansions gives \(3 + \dfrac{25}{3}x + \dfrac{289}{9}x^2\). For positive \(x\) every term of both expansions is positive, so the terms left out add a positive amount and the three-term approximation is too small: it is an underestimate. Common errors are writing \(3(3 - x)^{-1}\) as \(3\left(1 - \dfrac{x}{3}\right)^{-1}\), giving decimal coefficients and justifying the answer only by substituting a value of \(x\).

Mark scheme

1 mark
M1 A correct method for the partial fractions, e.g. \(9 - 14x \equiv A(3 - x) + B(1 - 4x)\) with \(x = \dfrac{1}{4}\) or \(x = 3\) substituted, or comparing coefficients; or correct final answers with no working (3 marks, see comments)
An answer of \(\dfrac{2}{1 - 4x} + \dfrac{3}{3 - x}\) and \(3 + \dfrac{25}{3}x + \dfrac{289}{9}x^2\) with no working scores 3 marks: the M1 A1 for the partial fractions and the final A1 for the expansion as a special case; the B1, M1 and the next two A1 marks need the expansions to be seen, and the last mark needs a reason. An identity with the brackets the wrong way round, \(9 - 14x \equiv A(1 - 4x) + B(3 - x)\) for \(\dfrac{A}{1 - 4x} + \dfrac{B}{3 - x}\), scores M0.
1 mark
A1 \(\mathrm{f}(x) = \dfrac{2}{1 - 4x} + \dfrac{3}{3 - x}\)
The partial fractions must be written out: \(A = 2\), \(B = 3\) alone scores A0 here. Accept \(-\dfrac{3}{x - 3}\) for \(\dfrac{3}{3 - x}\).
1 mark
B1 Writes \(\dfrac{3}{3 - x} = \left(1 - \dfrac{x}{3}\right)^{-1}\) (or \(3 \times 3^{-1}\left(1 - \dfrac{x}{3}\right)^{-1}\))
Takes out the factor 3 correctly. A common error is \(3\left(1 - \dfrac{x}{3}\right)^{-1}\), which scores B0.
1 mark
M1 Expands \((1 - 4x)^{-1}\) or \(\left(1 - \dfrac{x}{3}\right)^{-1}\) with the correct structure for the \(x^2\) term, e.g. \(\dfrac{(-1)(-2)}{2}(-4x)^2\)
Recognising either as the sum of a geometric series (common ratio \(4x\) or \(\dfrac{x}{3}\)) is equally good. Condone missing brackets if they are recovered.
1 mark
A1 One of the two expansions correct: \(2(1 - 4x)^{-1} = 2 + 8x + 32x^2 + \dots\) or \(\left(1 - \dfrac{x}{3}\right)^{-1} = 1 + \dfrac{1}{3}x + \dfrac{1}{9}x^2 + \dots\)
1 mark
A1 Both expansions correct
Depends on the M mark.
1 mark
A1 \(\mathrm{f}(x) = 3 + \dfrac{25}{3}x + \dfrac{289}{9}x^2 + \dots\)
Depends on the M mark. Accept \(8\tfrac{1}{3}x\) and \(32\tfrac{1}{9}x^2\). Rounded decimal coefficients such as \(8.33x\) score A0 (recurring decimals clearly indicated are acceptable). Ignore extra terms.
1 mark
B1 Underestimate, with a correct reason: for \(0 < x < \dfrac{1}{4}\) every term after the \(x^2\) term in both expansions is positive (e.g. the next terms are \(128x^3\) and \(\dfrac{1}{27}x^3\); both are geometric series with positive common ratios \(4x\) and \(\dfrac{x}{3}\)), so the terms left out have a positive sum
Both 'underestimate' and the reason are needed: 'underestimate' with no reason scores B0. Reasons based only on substituting a value of \(x\) (e.g. comparing with \(\mathrm{f}(0.1)\)), and vague statements such as 'more terms would be more accurate', score B0.

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