| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Factor condition (zero remainder) |
| Difficulty | Easy -1.2 This is a straightforward question testing understanding of the Factor Theorem at AS level. It requires recognizing that the student should have evaluated f(2) not f(-2), and that finding f(-2)=0 means (x+2) is a factor, not (x-2). No calculation or problem-solving is needed—just conceptual understanding of a basic theorem, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| 5(a) | States the student should have |
| Answer | Marks | Guidance |
|---|---|---|
| have calculated f(–2) | 2.3 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 5(b) | Infers that (x – 2) may or may |
| Answer | Marks | Guidance |
|---|---|---|
| factor | 2.2b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Deduces that (x + 2) is a factor | 2.2a | E1 |
| Subtotal | 2 | |
| Question 5 Total | 3 | |
| Q | Marking instructions | AO |
Question 5:
--- 5(a) ---
5(a) | States the student should have
calculated f(2)
or
States the student should not
have calculated f(–2) | 2.3 | E1 | The student should have calculated
f(2)
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 5(b) ---
5(b) | Infers that (x – 2) may or may
not be a factor
Do not accept definitive
conclusion that (x – 2) is not a
factor | 2.2b | E1 | They do not know whether (x – 2) is a
factor or not.
They could conclude that (x + 2) is a
factor.
Deduces that (x + 2) is a factor | 2.2a | E1
Subtotal | 2
Question 5 Total | 3
Q | Marking instructions | AO | Marks | Typical solutionA student is looking for factors of the polynomial $f(x)$
They suggest that $(x - 2)$ is a factor of $f(x)$
The method they use to check this suggestion is to calculate $f(-2)$
They correctly calculate that $f(-2) = 0$
They conclude that their suggestion is correct.
\begin{enumerate}[label=(\alph*)]
\item Make one comment about the student's method.
[1 mark]
\item Make two comments about the student's conclusion.
[2 marks]
1
2
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q5 [3]}}