| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.3 This is a straightforward application of the factor and remainder theorems requiring substitution into p(-2)=0 and p(3)=35 to form two simultaneous equations in a and b, followed by routine factorisation and analysis of roots. While it involves multiple steps (5+3 marks), each step uses standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | Substitute x=−2 and equate to zero | *M1 |
| Substitute x=3 and equate to 35 | *M1 | |
| Obtain −8a−4a−2a+b=0 and 27a−9a+3a+b=35 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| b | DM1 | Dependent at least one M mark. |
| Obtain a=1 and b=14 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(b) | Divide by x+2 at least as far as the x term | M1 |
| Obtain (x+2) (x2−3x+7) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| hence no root | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | Substitute x=−2 and equate to zero | *M1
Substitute x=3 and equate to 35 | *M1
Obtain −8a−4a−2a+b=0 and 27a−9a+3a+b=35 | A1
Solve a pair of relevant simultaneous linear equations to find a or
b | DM1 | Dependent at least one M mark.
Obtain a=1 and b=14 | A1
5
--- 3(b) ---
3(b) | Divide by x+2 at least as far as the x term | M1
Obtain (x+2) (x2−3x+7) | A1
Conclude with reference to −2, and discriminant is 9−28 and
hence no root | A1 | OE
3
Question | Answer | Marks | GuidanceThe polynomial $p(x)$ is defined by
$$p(x) = ax^3 - ax^2 + ax + b,$$
where $a$ and $b$ are constants. It is given that $(x + 2)$ is a factor of $p(x)$, and that the remainder is 35 when $p(x)$ is divided by $(x - 3)$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$. [5]
\item Hence factorise $p(x)$ and show that the equation $p(x) = 0$ has exactly one real root. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q3 [8]}}