| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single unknown constant |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor Theorem requiring substitution of x=-1 to find a constant, followed by polynomial division or substitution to find a remainder. Both parts are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial since it involves algebraic manipulation across two parts. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(x = -1\) and equate to zero | M1 | |
| Obtain answer \(a = 7\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(x = -3\) and evaluate expression | M1 | |
| Obtain answer 18 | A1 | |
| Total | 2 |
## Question 1:
**Part 1(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -1$ and equate to zero | M1 | |
| Obtain answer $a = 7$ | A1 | |
| **Total** | **2** | |
**Part 1(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -3$ and evaluate expression | M1 | |
| Obtain answer 18 | A1 | |
| **Total** | **2** | |
---1 (a) The polynomial $2 x ^ { 3 } + a x ^ { 2 } - a x - 12$, where $a$ is a constant, is denoted by $\mathrm { p } ( x )$. It is given that $( x + 1 )$ is a factor of $\mathrm { p } ( x )$.
Find the value of $a$.\\
(b) When $a$ has this value, find the remainder when $\mathrm { p } ( x )$ is divided by $( x + 3 )$.\\
\hfill \mbox{\textit{CAIE P2 2020 Q1 [4]}}