AQA GCSE Mathematics Paper 3 (Higher), June 2025: Question 20

Quadratics: Non-Monic Factorisation & Prime Sequences

📋 What this question tests

This question assesses higher-tier algebraic manipulation and deductive reasoning:

Factorising a non-monic quadratic: Splitting the middle term of an expression of the form an² + bn + c where a ≠ 1 .
Algebraic reasoning with sequences: Connecting the factorised form to the mathematical definition of a prime number (terms in a sequence have integer positions n ≥ 1 ).

Question 20 (a)

Factorise fully: 3n² + 5n + 2 [2 marks]

📐 Step-by-Step Factorisation

Identify coefficients:
For 3n² + 5n + 2 , a = 3 , b = 5 , c = 2 .
Find two numbers:
They must multiply to a × c = 3 × 2 = 6 and add to b = 5 .
The pair is 2 and 3.
Split the middle term:
3n² + 3n + 2n + 2
Factorise in pairs:
3n(n + 1) + 2(n + 1)
Extract the common bracket:
(3n + 2)(n + 1)

✅ Correct Answer & Marks

(3n + 2)(n + 1) (or (n + 1)(3n + 2) )

Mark Breakdown:

B2: Fully correct factorised expression.
B1: Partial progress: one correct factor seen in a product of two linear brackets, e.g. (3n + 2)(...) or (...)(n + 1) , or partially factorised as 3n(n + 1) + 2(n + 1) .

🧠 Exam Technique

Always expand to check: Multiply out (3n + 2)(n + 1) = 3n² + 3n + 2n + 2 = 3n² + 5n + 2 to guarantee your 2 marks.
Ignore solving: If you accidentally write n = -2/3, -1 , examiners will condone it and award full marks if the brackets are shown, but do not waste precious exam time solving when the question says "factorise".

❌ Common Errors

Wrong pair of factors: Using 1 and 6 because 6 - 1 = 5 , ignoring the sign requirements ( +6 product).
Writing as an equation: Writing (3n + 2)(n + 1) = 0 without leaving the factorised expression clearly identifiable.
Sign slips: Writing (3n - 2)(n - 1) .

Question 20 (b)

A sequence has nth term 3n² + 5n + 2. Are any of the terms in the sequence a prime number? Tick a box. Give a reason for your answer. [1 mark]

✅ Correct Answer

Box ticked: No [✓]

Acceptable Valid Reasons:

Factor argument (Best method):
Using part (a), the n th term is (3n + 2)(n + 1) . For a sequence, n ≥ 1 , so both factors are integers strictly greater than 1 ( n + 1 ≥ 2 and 3n + 2 ≥ 5 ). Since each term is a product of two integers greater than 1, it cannot be prime.
Parity argument:
Every term in the sequence is even, and the first term is 10 (which is greater than 2). Since 2 is the only even prime, no terms are prime.

Mark Breakdown:

B1: Ticking No AND giving a valid, fully justified mathematical reason.

💡 Key Knowledge

Sequences use positive integers: The position in a sequence is always n = 1, 2, 3, ... (natural numbers).
Definition of a Prime Number: A whole number strictly greater than 1 with exactly two distinct factors: 1 and itself.
Composite by Factorisation: If an integer N = A × B where A > 1 and B > 1 , then N is composite, never prime.
The "Even" Trap: 2 is an even number AND a prime number!

🧠 Exam Technique: Structuring the Reason

Questions with an algebraic part (a) usually link directly to part (b)!

Look back at your answer to part (a): (3n + 2)(n + 1) .
Ask yourself: Can either bracket ever equal 1 if n is 1, 2, 3...?
Smallest values occur at n = 1 : 3(1) + 2 = 5 and 1 + 1 = 2 .
State clearly: "The terms are the product of two factors, neither of which is equal to 1."

❌ Examiner Pitfalls & Insufficient Reasons

Incomplete even argument (B0): Stating only "Every term is even" scores 0 marks because the number 2 is even and prime! You must state that all terms are even and greater than 2 (or that the first term is 10).
False generalisations (B0): Writing "Adding 2 means it can't be prime" or "3n² can never be prime" gets no credit.
Only testing numbers (B0): Calculating the first few terms (10, 24, 44...) and saying "None of the ones I calculated are prime" is not a general proof!