| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Known polynomial, verify then factorise |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard A-level techniques: factor theorem application (routine substitution), polynomial division/comparison of coefficients (mechanical process), and a simple logical deduction about factorization. All parts follow predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| 13(a) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 4 2 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 2.1 | R1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 13(b) | Obtains two correct coefficients | |
| of 2 x 2 + 3 x + 4 | 1.1a | M1 |
| Obtains ( 2 x + 1 ) ( 2 x 2 + 3 x + 4 ) | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 13(c) | Begins argument by explaining |
| Answer | Marks | Guidance |
|---|---|---|
| Condone x instead of n | 2.1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 n 3 + 8 n 2 + 1 1 n + 4 | 2.2a | R1F |
| Subtotal | 2 | |
| Question 13 Total | 6 | |
| Q | Marking instructions | AO |
Question 13:
--- 13(a) ---
13(a) | 1
Substitutes x = − into P ( x )
2
and obtains zero.
1
Must see − bracketed
2
correctly.
If bracket(s) missing must see a
further step to indicate correct
evaluation eg
4 8 11
− + − +4=0or better.
8 4 2 | 1.1a | M1 | 3 2
1 1 1 1
P − = 4 − + 8 − + 1 1 − + 4
2 2 2 2
= 0
( 2 x + 1 ) is a factor of P ( x )
Completes factor theorem
argument by showing
1
P − =0 and stating
2
( 2 x + 1 ) is a factor of P(x)
OE | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 13(b) ---
13(b) | Obtains two correct coefficients
of 2 x 2 + 3 x + 4 | 1.1a | M1 | P(x)=( 2x+1 )( 2x2 +3x+4 )
Obtains ( 2 x + 1 ) ( 2 x 2 + 3 x + 4 ) | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 13(c) ---
13(c) | Begins argument by explaining
that either
( 2 n + 1 ) 1
Or
( )
an2 +bn+c 1
Or
( 2 n + 1 ) the cubic expression
Or
( )
an2 +bn+c the cubic
expression
Condone x instead of n | 2.1 | M1 | 4 n 3 + 8 n 2 + 1 1 n + 4 = ( 2 n + 1 ) ( 2 n 2 + 3 n + 4 )
There are two factors. Both factors are
integers not equal to 1 so
4 n 3 + 8 n 2 + 1 1 n + 4 is never prime.
States that
Either
both ( 2 n + 1 ) 1 and their
( )
an2 +bn+c 1
Or
both ( 2 n + 1 ) is not equal to
4n3 +8n2 +11n+4
( )
and their a n 2 + b n + c is not
equal to 4 n 3 + 8 n 2 + 1 1 n + 4
Or
( 2 n + 1 ) 1 and ( 2 n + 1 ) is not
equal to 4 n 3 + 8 n 2 + 1 1 n + 4
Or
( )
their a n 2 + b n + c 1 and their
( )
an2 +bn+c is not equal to
4 n 3 + 8 n 2 + 1 1 n + 4 | 2.2a | R1F
Subtotal | 2
Question 13 Total | 6
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item It is given that
$$P(x) = 4x^3 + 8x^2 + 11x + 4$$
Use the factor theorem to show that $(2x + 1)$ is a factor of $P(x)$
[2 marks]
\item Express $P(x)$ in the form
$$P(x) = (2x + 1)(ax^2 + bx + c)$$
where $a$, $b$ and $c$ are constants to be found.
[2 marks]
\item Given that $n$ is a positive integer, use your answer to part (b) to explain why $4n^3 + 8n^2 + 11n + 4$ is never prime.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2024 Q13 [6]}}