| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Sketching Polynomial Curves |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C1 techniques: factor theorem with trial-and-error to find roots, polynomial factorisation, curve sketching from factored form, and translation of polynomials. All parts are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-1)\) substituted for \(x\) in either form of equation for \(y = f(x)\) | M1 | correct or ft their (i) or (ii) for factorised form; condone one error; allow for new roots stated as \(-4, -2\) and \(3\) or ft |
| \((x-1)^3\) expanded correctly (need not be simplified) or two of their factors multiplied correctly | M1 dep | or M1 for correct or correct ft multiplying out of all 3 brackets at once, condoning one error \([x^3 - 3x^2 + x^2 + 2x^2 + 8x - 6x - 12x - 24]\) |
| correct completion to given answer [condone omission of '\(y=\)'] | M1 | unless all 3 brackets already expanded, must show at least one further interim step; allow SC1 for \((x+1)\) subst and correct exp of \((x+1)^3\) or two of their factors ft |
| or, for those using given answer: M1 for roots stated or used as \(-4,-2\) and \(3\) or ft; A1 for showing all 3 roots satisfy given eqn; B1 for comment re coefft of \(x^3\) or product of roots to show that eqn of translated graph is not a multiple of RHS of given eqn |
## Question 1(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-1)$ substituted for $x$ in either form of equation for $y = f(x)$ | M1 | correct or ft their (i) or (ii) for factorised form; condone one error; allow for new roots stated as $-4, -2$ and $3$ or ft |
| $(x-1)^3$ expanded correctly (need not be simplified) or two of their factors multiplied correctly | M1 dep | or M1 for correct or correct ft multiplying out of all 3 brackets at once, condoning one error $[x^3 - 3x^2 + x^2 + 2x^2 + 8x - 6x - 12x - 24]$ |
| correct completion to given answer [condone omission of '$y=$'] | M1 | unless all 3 brackets already expanded, must show at least one further interim step; allow SC1 for $(x+1)$ subst and correct exp of $(x+1)^3$ or two of their factors ft |
| | | or, for those using given answer: M1 for roots stated or used as $-4,-2$ and $3$ or ft; A1 for showing all 3 roots satisfy given eqn; B1 for comment re coefft of $x^3$ or product of roots to show that eqn of translated graph is not a multiple of RHS of given eqn |
---1 You are given that $\mathrm { f } ( x ) = x ^ { 3 } + 6 x ^ { 2 } - x - 30$.\\
(i) Use the factor theorem to find a root of $\mathrm { f } ( x ) = 0$ and hence factorise $\mathrm { f } ( x )$ completely.\\
(ii) Sketch the graph of $y = \mathrm { f } ( x )$.\\
(iii) The graph of $y = \mathrm { f } ( x )$ is translated by $\binom { 1 } { 0 }$.
Show that the equation of the translated graph may be written as
$$y = x ^ { 3 } + 3 x ^ { 2 } - 10 x - 24$$
\hfill \mbox{\textit{OCR MEI C1 Q1 [12]}}