| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify factor then simplify rational expression |
| Difficulty | Moderate -0.3 Part (a) is straightforward application of the factor theorem requiring only substitution of x=-3. Part (b) requires factorising the cubic using the result from (a), factorising the denominator as difference of two squares, then cancelling common factors—this is a standard AS-level algebraic manipulation exercise with clear scaffolding from part (a), making it slightly easier than average overall. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | Demonstrates p(3)0 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| this implies that x + 3 is a factor) | AO2.1 | R1 |
| (b) | Factorises the numerator and |
| Answer | Marks | Guidance |
|---|---|---|
| division) | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| complete factorisation) | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| factorisation) | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| stated | AO1.1b | A1 |
| Total | 6 | |
| Q | Marking Instructions | AO |
Question 4:
--- 4(a) ---
4(a) | Demonstrates p(3)0 | AO1.1b | B1 | p(3)2(3)3 7(3)2 2(3)3
546363 0
p(–3) = 0 x3 is a factor
Constructs rigorous mathematical
proof
(to achieve this mark, the student
must clearly calculate and state
that p(3)0and clearly state that
this implies that x + 3 is a factor) | AO2.1 | R1
(b) | Factorises the numerator and
denominator
(this mark is achieved for any
reasonable attempt at factorisation
through the selection of an
appropriate method, eg long
division) | AO1.1a | M1 | (x3)(2x2x1)
(2x1)(2x1)
(x3)(2x1)(x+1)
(2x1)(2x1)
(x3)(x1) 1
, x
(2x1) 2
Finds second factor in numerator or
fully factorises denominator (PI by
complete factorisation) | AO1.1b | A1
Finds fully correct factorised
expression (PI by complete
factorisation) | AO1.1b | A1
Obtains a completely correct
solution with restriction on domain
stated | AO1.1b | A1
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution$p(x) = 2x^3 + 7x^2 + 2x - 3$
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to prove that $x + 3$ is a factor of $p(x)$
[2 marks]
\item Simplify the expression $\frac{2x^3 + 7x^2 + 2x - 3}{4x^2 - 1}$, $x \neq \pm \frac{1}{2}$
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 Q4 [6]}}