| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Division then Show Number of Real Roots |
| Difficulty | Standard +0.3 This is a straightforward polynomial division question with standard techniques. Part (i) is routine long division, part (ii) uses the remainder theorem to set up simple equations, and part (iii) requires factorizing and analyzing roots of a quadratic—all standard A-level procedures with no novel insight required. Slightly above average due to the multi-part nature and need to connect concepts across parts. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Carry out division at least as far as quotient \(x^2 + kx\) | M1 | |
| Obtain partial quotient \(x^2 + 2x\) | A1 | |
| Obtain quotient \(x^2 + 2x + 1\) with no errors seen | A1 | |
| Obtain remainder \(5x + 2\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either carry out calculation involving \(12x + 6\) and their remainder \(ax + b\), or multiply \(x^2 - x + 4\) by their three-term quadratic quotient | M1 | |
| Obtain \(p = 7, q = 4\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Show that discriminant of \(x^2 - x + 4\) is negative | B1 | |
| Form equation \((x^2 - x + 4)(x^2 + 2x + 1) = 0\) and attempt solution | M1 | |
| Show that \(x^2 + 2x + 1 = 0\) gives one root \(x = -1\) | A1 | |
| Total: 3 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Carry out division at least as far as quotient $x^2 + kx$ | M1 | |
| Obtain partial quotient $x^2 + 2x$ | A1 | |
| Obtain quotient $x^2 + 2x + 1$ with no errors seen | A1 | |
| Obtain remainder $5x + 2$ | A1 | |
| **Total: 4** | | |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Either carry out calculation involving $12x + 6$ and their remainder $ax + b$, or multiply $x^2 - x + 4$ by their three-term quadratic quotient | M1 | |
| Obtain $p = 7, q = 4$ | A1 | |
| **Total: 2** | | |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Show that discriminant of $x^2 - x + 4$ is negative | B1 | |
| Form equation $(x^2 - x + 4)(x^2 + 2x + 1) = 0$ and attempt solution | M1 | |
| Show that $x^2 + 2x + 1 = 0$ gives one root $x = -1$ | A1 | |
| **Total: 3** | | |
---6 (i) Find the quotient and remainder when
$$x ^ { 4 } + x ^ { 3 } + 3 x ^ { 2 } + 12 x + 6$$
is divided by ( $x ^ { 2 } - x + 4$ ).\\
(ii) It is given that, when
$$x ^ { 4 } + x ^ { 3 } + 3 x ^ { 2 } + p x + q$$
is divided by $\left( x ^ { 2 } - x + 4 \right)$, the remainder is zero. Find the values of the constants $p$ and $q$.\\
(iii) When $p$ and $q$ have these values, show that there is exactly one real value of $x$ satisfying the equation
$$x ^ { 4 } + x ^ { 3 } + 3 x ^ { 2 } + p x + q = 0$$
and state what that value is.\\
\hfill \mbox{\textit{CAIE P2 Q6 [9]}}