| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2003 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Polynomial Identity Matching |
| Difficulty | Moderate -0.3 This is a structured polynomial identity problem with clear guidance. Part (i) requires expanding the product and comparing coefficients—a routine algebraic technique. Part (ii) uses the factorization to solve two quadratics via the quadratic formula. While it involves multiple steps and algebraic manipulation, the path is clearly signposted and uses standard A-level methods without requiring insight or problem-solving creativity. |
| 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 |
| EITHER: Expand RHS and obtain at least one equation for \(a\) | M1 | |
| Obtain \(a^2 = 9\) and \(2a = 6\), or equivalent | A1 | |
| State answer \(a = 3\) only | A1 | |
| OR: Attempt division by \(x^2 + ax + 1\) or \(x^2 - ax - 1\), and obtain an equation in \(a\) | M1 | |
| Obtain \(a^2 = 9\) and either \(a^3 - 11a + 6 = 0\) or \(a^3 - 7a - 6 = 0\), or equivalent | A1 | |
| State answer \(a = 3\) only | A1 | |
| Special case: the answer \(a = 3\), obtained by trial and error, or by inspection, or with no working | B2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute for \(a\) and attempt to find zeroes of one of the quadratic factors | M1 | |
| Obtain one correct answer | A1 | |
| State all four solutions \(\frac{1}{2}(-3 \pm \sqrt{5})\) and \(\frac{1}{2}(3 \pm \sqrt{13})\), or equivalent | A1 |
# Question 2(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER:** Expand RHS and obtain at least one equation for $a$ | M1 | |
| Obtain $a^2 = 9$ and $2a = 6$, or equivalent | A1 | |
| State answer $a = 3$ only | A1 | |
| **OR:** Attempt division by $x^2 + ax + 1$ or $x^2 - ax - 1$, and obtain an equation in $a$ | M1 | |
| Obtain $a^2 = 9$ and either $a^3 - 11a + 6 = 0$ or $a^3 - 7a - 6 = 0$, or equivalent | A1 | |
| State answer $a = 3$ only | A1 | |
| Special case: the answer $a = 3$, obtained by trial and error, or by inspection, or with no working | | B2 |
**Total: [3]**
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# Question 2(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute for $a$ and attempt to find zeroes of one of the quadratic factors | M1 | |
| Obtain one correct answer | A1 | |
| State all four solutions $\frac{1}{2}(-3 \pm \sqrt{5})$ and $\frac{1}{2}(3 \pm \sqrt{13})$, or equivalent | A1 | |
**Total: [3]**
---2 The polynomial $x ^ { 4 } - 9 x ^ { 2 } - 6 x - 1$ is denoted by $\mathrm { f } ( x )$.\\
(i) Find the value of the constant $a$ for which
$$f ( x ) \equiv \left( x ^ { 2 } + a x + 1 \right) \left( x ^ { 2 } - a x - 1 \right)$$
(ii) Hence solve the equation $\mathrm { f } ( x ) = 0$, giving your answers in an exact form.
\hfill \mbox{\textit{CAIE P2 2003 Q2 [6]}}