| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Fully specified polynomial: verify factor and solve |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring substitution to verify a factor, then polynomial division and solving a quadratic. It's routine C1 material with clear steps and no problem-solving insight needed, making it easier than average but not trivial since it requires multiple techniques. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(2) = 8 - 14 + 6 = 0\) so \((x-2)\) is a factor | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (x-2)(x^2+2x-3) = 0\) | M1 A1 | |
| \(\Rightarrow (x-2)(x-1)(x+3) = 0\) | A1 | |
| \(\Rightarrow x = 1, 2, -3\) | B1 | Complete factorisation. Total: 4 |
# Question 5:
## Part (i):
$f(2) = 8 - 14 + 6 = 0$ so $(x-2)$ is a factor | B1 | **Total: 1**
## Part (ii):
$f(x) = (x-2)(x^2+2x-3) = 0$ | M1 A1 |
$\Rightarrow (x-2)(x-1)(x+3) = 0$ | A1 |
$\Rightarrow x = 1, 2, -3$ | B1 | Complete factorisation. **Total: 4**
---5 You are given that $\mathrm { f } ( x ) = x ^ { 3 } - 7 x + 6$.\\
(i) Show that ( $x - 2$ ) is a factor of $\mathrm { f } ( x )$.\\
(ii) Solve the equation $\mathrm { f } ( x ) = 0$.
\hfill \mbox{\textit{OCR MEI C1 Q5 [5]}}