AQA AS Level Mathematics Paper 1, June 2025: Question 3

Mark scheme

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Q Marking instructions AO Marks Typical solution

3 Substitutes x = 2 into p(x) 1.1a M1 2 2( )3 − a( )22 + 6 2( ) + 2a = 0

Sets their p(2) equal to zero and 1.1a M1 − a + + a =

16 4 12 2 0

solves to obtain a value for a

PI by a = 14 28 − 2a = 0

Obtains a = 14 1.1b A1 a = 14

Question 3 Total 3

How to answer it

Applying the Factor Theorem to Find an Unknown Coefficient

📌 What this question tests

This question assesses your ability to apply the Factor Theorem to determine an unknown constant within a cubic polynomial equation:

Understanding that if (x - c) is a factor of a polynomial p(x) , then p(c) = 0 .
Correct algebraic substitution into terms with indices: evaluating x³ and x² .
Collecting like terms and solving a simple linear equation for an unknown coefficient a .

Question 3 • 3 Marks

Full Worked Solution & Mark Breakdown

Given: p(x) = 2x³ - ax² + 6x + 2a, with factor (x - 2)

📐 Step-by-Step Solution

1 State and apply the Factor Theorem:
Since (x - 2) is a factor of p(x) , it follows that p(2) = 0 .

2 Substitute x = 2 into p(x):
p(2) = 2(2)³ - a(2)² + 6(2) + 2a

3 Evaluate each term:
• 2(2)³ = 2(8) = 16
• -a(2)² = -4a
• 6(2) = 12
• +2a remains +2a
So: 16 - 4a + 12 + 2a = 0

4 Simplify and solve for a:
(16 + 12) + (-4a + 2a) = 0
28 - 2a = 0
2a = 28
a = 14

✅ Final Answer & Marks

a = 14

Mark Scheme Breakdown:
• M1 (AO 1.1a): Valid attempt to substitute x = 2 into p(x) .
• M1 (AO 1.1a): Sets their p(2) = 0 and attempts to solve for a (implied by obtaining a = 14 ).
• A1 (AO 1.1b): Correct final value a = 14 .

💡 Key Knowledge

The Factor Theorem: A polynomial p(x) has a linear factor (x - c) if and only if p(c) = 0 .
Roots vs. Factors: If the factor is (x - 2) , the root being substituted is x = +2 (the solution to x - 2 = 0 ).
Linear Equation Strategy: Group constant numerical terms together and unknown parameter terms together before rearranging.

🧠 Exam Technique

Show the substitution explicitly: Write p(2) = 2(2)³ - a(2)² + 6(2) + 2a = 0 clearly before simplifying. This secures the first method mark even if an arithmetic slip happens later.
Check by back-substitution: Substitute a = 14 back into p(2) :
2(8) - 14(4) + 6(2) + 2(14) = 16 - 56 + 12 + 28 = 0 . Zero confirms your answer is 100% correct!
Follow the question rubric: The question specifies "by using the factor theorem" — avoid polynomial long division here, as using the factor theorem is mandatory.

❌ Common Errors & Traps

Sign reversal on substitution: Substituting x = -2 instead of x = 2 . Remember: x - 2 = 0 ⇒ x = 2 .
Incorrect order of operations (BIDMAS): Miscalculating -a(2)² as -(2a)² = -4a² . Powers apply only to the base 2 , yielding -4a .
Algebraic collection slip: Combining -4a + 2a incorrectly as -6a or +2a instead of -2a .
Forgetting to equate to zero: Simplifying the expression to 28 - 2a but failing to form an equation ( = 0 ) to solve.

Topics

Pure Mathematics · B: Algebra and functions