Mathematics 9709 · syllabus 2.1, 3.1
Cambridge A Level Mathematics 9709 — Algebra topical past papers
The modulus function, polynomial division and the factor and remainder theorems, plus partial fractions and binomial series for any rational power in Paper 3.
- Questions
- 195
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 71
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What Algebra covers
Algebra in Papers 2 and 3 (sections 2.1 and 3.1) begins with the modulus function. You sketch y = |ax + b|, a V shape with the part below the x-axis reflected upwards, and solve equations and inequalities with moduli, either by squaring both sides or by taking cases, with a sketch to check. Then come polynomials: dividing a polynomial of degree up to 4 by a linear or quadratic divisor, and using the factor and remainder theorems to find factors, remainders and unknown coefficients.
Paper 3 adds two more ideas. Partial fractions split a rational expression with a denominator such as (ax + b)(cx + d)(ex + f), (ax + b)(cx + d)² or (ax + b)(cx² + d) into simpler fractions. The binomial expansion of (1 + x)ⁿ for any rational n gives an infinite series valid for |x| < 1, and adapting it to an expression like (4 − x)^(−½) means taking out a factor first and finding the new range of validity.
Paper 2 stops at the modulus function, polynomial division and the two theorems, so on that route the topic is shorter. In Paper 3, partial fractions make many integrals and differential equations possible, and combined with binomial series they let you expand awkward fractions as power series. The factor theorem is used again when solving cubic and quartic equations with complex roots, and the modulus reappears as a distance on an Argand diagram.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 2.1 Algebra
- 3.1 Algebra
What you should be able to do
- Sketch y = |ax + b| and solve modulus equations and inequalities
- Divide a polynomial by a linear or quadratic polynomial, finding the quotient and remainder
- Use the factor and remainder theorems to find unknown coefficients and to factorise fully
- Split a rational function into partial fractions, including repeated and quadratic factors (Paper 3)
- Expand (1 + x)ⁿ for rational n and state the values of x for which it is valid (Paper 3)
Key ideas
- |a| = |b| ⇔ a² = b², and |x − a| < b ⇔ a − b < x < a + b
- Remainder theorem: dividing p(x) by (ax + b) leaves remainder p(−b/a); factor theorem: (ax + b) is a factor if and only if p(−b/a) = 0
- (px + q)/((ax + b)(cx + d)²) = A/(ax + b) + B/(cx + d) + C/(cx + d)², and a factor (cx² + d) takes a numerator Bx + C
- (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + …, valid for |x| < 1 when n is not a positive integer
How Algebra has been examined
From 2022 to 2026, Algebra was on Paper 3 · Pure Mathematics 3 in 11 of 11 exam series, worth about 18% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 3 · Pure Mathematics 3 | 71 | 18% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 5.
Filed in Momentum under: Algebra.
A worked example
The polynomial p(x) = 2x³ + ax² − 7x + b, where a and b are constants, has (x − 2) as a factor and leaves a remainder of 6 when divided by (x + 1). Find a and b.
- (x − 2) is a factor, so p(2) = 0: 16 + 4a − 14 + b = 0, giving 4a + b = −2.
- The remainder on dividing by (x + 1) is p(−1): −2 + a + 7 + b = 6, giving a + b = 1.
- Subtract the second equation from the first: 3a = −3, so a = −1 and then b = 2.
- Check with p(x) = 2x³ − x² − 7x + 2: p(2) = 16 − 4 − 14 + 2 = 0 and p(−1) = −2 − 1 + 7 + 2 = 6.
Answer: a = −1, b = 2
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Squaring both sides of a modulus inequality when one side may be negative: squaring is only safe when both sides are moduli or both are non-negative, so otherwise solve by cases.
- Substituting x = 2 for division by (x + 2): the value to substitute is the root of the divisor, x = −2.
- Giving a repeated factor only one partial fraction: (x − 1)² needs both B/(x − 1) and C/(x − 1)².
- Expanding (4 − x)^(−½) as if it were (1 − x)^(−½): take out 4^(−½) = ½ first to get ½(1 − x/4)^(−½), which is valid for |x| < 4.
- Not checking solutions of a modulus equation found by cases, which can produce a value that does not satisfy the original equation.
Exam technique
- Sketch both sides of a modulus equation or inequality; the sketch shows how many solutions there are and which region you need.
- For partial fractions, substitute the roots of the denominator to find constants quickly, then compare coefficients for any that remain.
- Simplify each coefficient of a binomial series fully and state the validity condition, which is often a separate mark.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
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Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.