| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2004 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One unknown constant: find it then solve |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring substitution to find 'a', then factorisation and solving a quadratic. All steps are routine AS-level techniques with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x = 3\) and equate to zero | M1 | |
| Obtain answer \(\alpha = -1\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At any stage, state that \(x = 3\) is a solution | B1 | |
| EITHER: Attempt division by \((x-3)\) reaching a partial quotient of \(2x^2 + kx\) | M1 | |
| Obtain quadratic factor \(2x^2 + 5x + 2\) | A1 | |
| Obtain solutions \(x = -2\) and \(x = -\frac{1}{2}\) | A1 | |
| OR: Obtain solution \(x = -2\) by trial and error | B1 | |
| Obtain solution \(x = -\frac{1}{2}\) similarly | B2 | Total: 4 |
| If an attempt at the quadratic factor is made by inspection, the M1 is earned if it reaches an unknown factor of \(2x^2 + bx + c\) and an equation in b and/or c. |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = 3$ and equate to zero | M1 | |
| Obtain answer $\alpha = -1$ | A1 | **Total: 2** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At any stage, state that $x = 3$ is a solution | B1 | |
| EITHER: Attempt division by $(x-3)$ reaching a partial quotient of $2x^2 + kx$ | M1 | |
| Obtain quadratic factor $2x^2 + 5x + 2$ | A1 | |
| Obtain solutions $x = -2$ and $x = -\frac{1}{2}$ | A1 | |
| OR: Obtain solution $x = -2$ by trial and error | B1 | |
| Obtain solution $x = -\frac{1}{2}$ similarly | B2 | **Total: 4** |
| | | If an attempt at the quadratic factor is made by inspection, the M1 is earned if it reaches an unknown factor of $2x^2 + bx + c$ and an equation in b and/or c. |
---3 The cubic polynomial $2 x ^ { 3 } + a x ^ { 2 } - 13 x - 6$ is denoted by $\mathrm { f } ( x )$. It is given that ( $x - 3$ ) is a factor of $\mathrm { f } ( x )$.\\
(i) Find the value of $a$.\\
(ii) When $a$ has this value, solve the equation $\mathrm { f } ( x ) = 0$.
\hfill \mbox{\textit{CAIE P2 2004 Q3 [6]}}