| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2008 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constant then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring substitution to find the constant, followed by polynomial division and factorisation of a quadratic. It's routine bookwork with clear signposting and no problem-solving insight needed, making it easier than average but not trivial since it requires multiple standard techniques. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -2\) and equate result to zero, or divide by \(x + 2\) and equate constant remainder to zero | M1 | |
| Obtain answer \(a = -13\) | A1 | [2] |
| (ii) Obtain quadratic factor \(2x^2 - 5x - 3\) | B1 | |
| Obtain linear factor \(2x + 1\) | B1 | |
| Obtain linear factor \(x - 3\) | B1 | [3] |
**(i)** Substitute $x = -2$ and equate result to zero, or divide by $x + 2$ and equate constant remainder to zero | M1 |
Obtain answer $a = -13$ | A1 | [2]
**(ii)** Obtain quadratic factor $2x^2 - 5x - 3$ | B1 |
Obtain linear factor $2x + 1$ | B1 |
Obtain linear factor $x - 3$ | B1 | [3]
[Condone omission of repetition that $x + 2$ is a factor.]
[If linear factors $2x + 1, x - 3$ obtained by remainder theorem or inspection, award B2 + B1.]2 The polynomial $2 x ^ { 3 } - x ^ { 2 } + a x - 6$, where $a$ is a constant, is denoted by $\mathrm { p } ( x )$. It is given that ( $x + 2$ ) is a factor of $\mathrm { p } ( x )$.\\
(i) Find the value of $a$.\\
(ii) When $a$ has this value, factorise $\mathrm { p } ( x )$ completely.
\hfill \mbox{\textit{CAIE P2 2008 Q2 [5]}}