| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Factorisation After Division or Remainder |
| Difficulty | Standard +0.3 This is a straightforward polynomial division question requiring standard algebraic manipulation. Part (i) involves routine long division or comparison of coefficients, while part (ii) uses the result to factor and solve a quartic equation. The 'hence' structure guides students clearly through the method, making this slightly easier than average but still requiring careful execution across multiple steps. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Divide at least as far as \(x\) term in quotient, use synthetic division correctly or make use of an identity | M1 | |
| Obtain at least \(6x^2 - x\) | A1 | |
| Obtain quotient \(6x^2 - x - 2\) and confirm remainder is \(7\) (AG) | A1 | [3] |
| (ii) State equation in form \(\left(x^2 - 4\right)\left(6x^2 + kx - 2\right) = 0\), any constant \(k\) (may be implied) | M1 | |
| Obtain two of the roots \(-2, 2, -\frac{1}{2}, \frac{2}{3}\) | A1 | |
| Obtain remaining two roots and no others | A1 | [3] |
**(i)** Divide at least as far as $x$ term in quotient, use synthetic division correctly or make use of an identity | M1 |
Obtain at least $6x^2 - x$ | A1 |
Obtain quotient $6x^2 - x - 2$ and confirm remainder is $7$ (AG) | A1 | [3]
**(ii)** State equation in form $\left(x^2 - 4\right)\left(6x^2 + kx - 2\right) = 0$, any constant $k$ (may be implied) | M1 |
Obtain two of the roots $-2, 2, -\frac{1}{2}, \frac{2}{3}$ | A1 |
Obtain remaining two roots and no others | A1 | [3]3 (i) Find the quotient when $6 x ^ { 4 } - x ^ { 3 } - 26 x ^ { 2 } + 4 x + 15$ is divided by ( $x ^ { 2 } - 4$ ), and confirm that the remainder is 7 .\\
(ii) Hence solve the equation $6 x ^ { 4 } - x ^ { 3 } - 26 x ^ { 2 } + 4 x + 8 = 0$.
\hfill \mbox{\textit{CAIE P2 2014 Q3 [6]}}