| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 6 |
| Topic | Factor & Remainder Theorem |
| Type | Find constants from coefficient conditions |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem with roots clearly visible from a graph. Students read off three roots, substitute into the general cubic form, and solve a simple system of linear equations. The multi-part structure guides students through each step with minimal problem-solving required beyond routine algebraic manipulation. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks |
|---|---|
| \(-4\) | B1 |
| Answer | Marks |
|---|---|
| \(x, (x + 2), (x - 1)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-4 = a(-1)(-2)(+1) \Rightarrow a = -2\) | M1, M1, A1if | ft their (i) and (ii) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -2x^3 - 2x^2 + 4x\) or \(b = -2, c = 4, d = 0\) | B1ft, M1 | ft their (ii) |
| [1] |
## 1(i)
$-4$ | B1 |
## 1(ii)
$x, (x + 2), (x - 1)$ | B1 |
## 1(iii)
$y = a(x(x - 1))(x + 2)$
Subst $(-1, -4)$ or from (i)
$-4 = a(-1)(-2)(+1) \Rightarrow a = -2$ | M1, M1, A1if | ft their (i) and (ii)
| [3]
## 1(iv)
$y = -2x(x - 1)(x + 2)$
$y = -2x^3 - 2x^2 + 4x$ or $b = -2, c = 4, d = 0$ | B1ft, M1 | ft their (ii)
| [1]
---1 Part of the graph of $y = \mathrm { f } ( x )$ is shown below, where $\mathrm { f } ( x )$ is a cubic polynomial.\\
\includegraphics[max width=\textwidth, alt={}, center]{6a6316e4-7b2d-4533-988a-4863d79ce668-04_681_679_475_694}\\
(i) Find $\mathrm { f } ( - 1 )$.\\
(ii) Write down three linear factors of $\mathrm { f } ( x )$.
It is given that $\mathrm { f } ( x ) \equiv a x ^ { 3 } + b x ^ { 2 } + c x + d$.\\
(iii) Show that $a = - 2$.\\
(iv) Find $b , c$ and $d$.
\hfill \mbox{\textit{OCR H240/02 2018 Q1 [6]}}