| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2006 |
| Session | November |
| 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 students to substitute x=3/2 to find a, then factorise and solve a cubic. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = \frac{2}{3}\) and equate to zero | M1 | |
| Obtain answer \(a = -3\) | A1 | 2 marks |
| (ii) At any stage, state that \(x = -\frac{2}{3}\) is a solution | B1 | |
| EITHER: Attempt division by \(2x - 3\) reaching a partial quotient of \(2x^2 + kx\) | M1 | |
| Obtain quadratic factor \(2x^2 + 3x + 1\) | A1 | |
| Obtain solutions \(x = -1\) and \(x = -\frac{1}{2}\) | A1 | |
| OR: Obtain solution \(x = -1\) by trial and error or inspection | B1 | |
| Obtain solution \(x = -\frac{1}{2}\) similarly | B2 | 4 marks |
(i) Substitute $x = \frac{2}{3}$ and equate to zero | M1 |
Obtain answer $a = -3$ | A1 | 2 marks |
(ii) At any stage, state that $x = -\frac{2}{3}$ is a solution | B1 |
EITHER: Attempt division by $2x - 3$ reaching a partial quotient of $2x^2 + kx$ | M1 |
Obtain quadratic factor $2x^2 + 3x + 1$ | A1 |
Obtain solutions $x = -1$ and $x = -\frac{1}{2}$ | A1 |
OR: Obtain solution $x = -1$ by trial and error or inspection | B1 |
Obtain solution $x = -\frac{1}{2}$ similarly | B2 | 4 marks |
[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 polynomial $4 x ^ { 3 } - 7 x + a$, where $a$ is a constant, is denoted by $\mathrm { p } ( x )$. It is given that ( $2 x - 3$ ) is a factor of $\mathrm { p } ( x )$.\\
(i) Show that $a = - 3$.\\
(ii) Hence, or otherwise, solve the equation $\mathrm { p } ( x ) = 0$.
\hfill \mbox{\textit{CAIE P2 2006 Q3 [6]}}