| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Prove root count with given polynomial |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic factor theorem application, polynomial division/expansion, and discriminant analysis. All parts follow standard textbook procedures with no problem-solving insight required—part (a) is simple substitution, part (b) is routine algebraic manipulation, and part (c) uses the discriminant formula to show no other real roots exist. Easier than average for AS-level. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(3) = 3^3 - 4\times3^2 + 10\times3 - 21 = 27 - 36 + 30 - 21 = 0\) | M1 | Substitutes \(x = 3\) into expression for \(f(x)\). Do not allow for algebraic division |
| [So by the factor theorem] \((x-3)\) is a factor | A1 [2] | Argues from zero clearly seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Algebraic division gives \(f(x) = (x-3)(x^2 - x + 7)\) | M1 | Attempts to divide by \((x-3)\) as far as linear term. Allow arithmetic slips. Also allow for expanding and equating coefficients |
| So \(b = -1\) and \(c = 7\) | A1 [2] | Allow seen in correct product of factors or correct algebraic division or multiplication grid. Correct factorisation by inspection scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{f}(x) = 0\) when \(x = 3]\) or when \(x^2 - x + 7 = 0\); Discriminant \((-1)^2 - 4\times3\times7 = -27 < 0\) | M1 | Finds discriminant. Also allow for equivalent argument using quadratic formula or completing the square |
| So no additional real roots | A1 [2] | www Clear argument from their negative discriminant. FT their \(b\) and \(c\). Condone missing reference to the fact that \(x = 3\) is a root |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(3) = 3^3 - 4\times3^2 + 10\times3 - 21 = 27 - 36 + 30 - 21 = 0$ | M1 | Substitutes $x = 3$ into expression for $f(x)$. Do not allow for algebraic division |
| [So by the factor theorem] $(x-3)$ is a factor | A1 [2] | Argues from zero clearly seen |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Algebraic division gives $f(x) = (x-3)(x^2 - x + 7)$ | M1 | Attempts to divide by $(x-3)$ as far as linear term. Allow arithmetic slips. Also allow for expanding and equating coefficients |
| So $b = -1$ and $c = 7$ | A1 [2] | Allow seen in correct product of factors or correct algebraic division or multiplication grid. Correct factorisation by inspection scores both marks |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{f}(x) = 0$ when $x = 3]$ or when $x^2 - x + 7 = 0$; Discriminant $(-1)^2 - 4\times3\times7 = -27 < 0$ | M1 | Finds discriminant. Also allow for equivalent argument using quadratic formula or completing the square |
| So no additional real roots | A1 [2] | www Clear argument from their negative discriminant. FT their $b$ and $c$. Condone missing reference to the fact that $x = 3$ is a root |
---6 The polynomial $x ^ { 3 } - 4 x ^ { 2 } + 10 x - 21$ is denoted by $\mathrm { f } ( x )$.
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $( x - 3 )$ is a factor of $\mathrm { f } ( x )$.
\item The polynomial $\mathrm { f } ( x )$ can be written as $( \mathrm { x } - 3 ) \left( \mathrm { x } ^ { 2 } + \mathrm { bx } + \mathrm { c } \right)$ where $b$ and $c$ are constants. Find the values of $b$ and $c$.
\item Show that $x = 3$ is the only real root of the equation $\mathrm { f } ( x ) = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q6 [6]}}