| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Known polynomial, verify then factorise |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on the Factor Theorem requiring standard techniques: identifying Nigel's error (should substitute x=-7, not x=7), verifying the factor, completing the factorisation, sketching a cubic with given intercepts, and applying a horizontal translation. All parts are routine AS-level procedures 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 |
| Nigel should substitute \(x = -7\) not \(x = 7\); his calculation shows that \((x-7)\) is not a factor | B1 | 2.3 – Any suitable comment that references the factor theorem. Allow for correct algebraic division if comment made about no remainder |
| \((-7)^3 - 37\times(-7) + 84 = -343 + 259 + 84 = 0\), so \((x+7)\) is a factor | B1 [2] | 2.1 – Clear argument needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 - 37x + 84 = (x+7)(x^2-7x+12)\) | M1 | 3.1a – Attempt to divide cubic by \((x+7)\) or find a quadratic factor by inspection. May be seen above (allow if at least two correct terms) |
| \(= (x+7)(x-3)(x-4)\) | M1 | 1.1a – Attempt to factorise their quadratic factor oe and find points on \(x\)-axis. FT |
| Crosses \(x\)-axis at \((-7,\ 0),\ (3,\ 0),\ (4,\ 0)\) | A1 (dep) | 1.1 – Accept values shown on sketch graph www. The A mark is dependent on the second M mark |
| Crosses \(y\)-axis at \((0,\ 84)\) | B1 | 1.1 – Accept value shown on sketch graph www. Exact position of stationary points is not needed |
| Axes labelled and correct general shape FT their values, right way up | B1 [5] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of the form \(y = (x-1)^3 - 37(x-1) + 84\) | M1 | 3.1a: Substituting \((x-1)\) for \(x\) twice |
| \(= (x^3 - 3x^2 + 3x - 1) - 37(x-1) + 84\) | M1 | 1.1a: Attempt to expand |
| Correct expansion of cubic term | A1 | 1.1: Correct expansion of their cubic term |
| \(y = x^3 - 3x^2 - 34x + 120\) | A1 [4] | 1.1: cao (including \(y=\)) |
| Alternative: \(y = (x-1+7)(x-1-3)(x-1-4)\) | M1 | Substituting \((x-1)\) for \(x\) everywhere; allow \((x+6)(x-4)(x-5)\) |
| \(= (x+6)(x^2 - 9x + 20)\) or \((x-4)(x^2 + x - 30)\) or \((x-5)(x^2 + 2x - 24)\) | M1, A1 | Attempt to multiply out their factors; correct quadratic factor |
| \(y = x^3 - 3x^2 - 34x + 120\) | A1 [4] | cao (including \(y=\)) |
## Question 7:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Nigel should substitute $x = -7$ not $x = 7$; his calculation shows that $(x-7)$ is not a factor | B1 | 2.3 – Any suitable comment that references the factor theorem. Allow for correct algebraic division if comment made about no remainder |
| $(-7)^3 - 37\times(-7) + 84 = -343 + 259 + 84 = 0$, so $(x+7)$ is a factor | B1 [2] | 2.1 – Clear argument needed |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 - 37x + 84 = (x+7)(x^2-7x+12)$ | M1 | 3.1a – Attempt to divide cubic by $(x+7)$ or find a quadratic factor by inspection. May be seen above (allow if at least two correct terms) |
| $= (x+7)(x-3)(x-4)$ | M1 | 1.1a – Attempt to factorise their quadratic factor oe and find points on $x$-axis. FT |
| Crosses $x$-axis at $(-7,\ 0),\ (3,\ 0),\ (4,\ 0)$ | A1 (dep) | 1.1 – Accept values shown on sketch graph www. The A mark is dependent on the second M mark |
| Crosses $y$-axis at $(0,\ 84)$ | B1 | 1.1 – Accept value shown on sketch graph www. Exact position of stationary points is not needed |
| Axes labelled and correct general shape FT their values, right way up | B1 [5] | 1.1 |
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of the form $y = (x-1)^3 - 37(x-1) + 84$ | M1 | 3.1a: Substituting $(x-1)$ for $x$ twice |
| $= (x^3 - 3x^2 + 3x - 1) - 37(x-1) + 84$ | M1 | 1.1a: Attempt to expand |
| Correct expansion of cubic term | A1 | 1.1: Correct expansion of their cubic term |
| $y = x^3 - 3x^2 - 34x + 120$ | A1 [4] | 1.1: cao (including $y=$) |
| **Alternative:** $y = (x-1+7)(x-1-3)(x-1-4)$ | M1 | Substituting $(x-1)$ for $x$ everywhere; allow $(x+6)(x-4)(x-5)$ |
| $= (x+6)(x^2 - 9x + 20)$ or $(x-4)(x^2 + x - 30)$ or $(x-5)(x^2 + 2x - 24)$ | M1, A1 | Attempt to multiply out their factors; correct quadratic factor |
| $y = x^3 - 3x^2 - 34x + 120$ | A1 [4] | cao (including $y=$) |
---7 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Nigel is asked to determine whether $( x + 7 )$ is a factor of $x ^ { 3 } - 37 x + 84$. He substitutes $x = 7$ and calculates $7 ^ { 3 } - 37 \times 7 + 84$. This comes to 168 , so Nigel concludes that ( $x + 7$ ) is not a factor.
Nigel's conclusion is wrong.
\begin{itemize}
\item Explain why Nigel's argument is not valid.
\item Show that $( x + 7 )$ is a factor of $x ^ { 3 } - 37 x + 84$.
\item Sketch the graph of $y = x ^ { 3 } - 37 x + 84$, indicating the coordinates of the points at which the curve crosses the coordinate axes.
\item The graph in part (b) is translated by $\binom { 1 } { 0 }$. Find the equation of the translated graph, giving your answer in the form $y = x ^ { 3 } + a x ^ { 2 } + b x + c$ where $a , b$ and $c$ are integers.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q7 [11]}}