| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| 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 C1 question testing basic polynomial manipulation, remainder theorem application, and discriminant analysis. Part (a) is routine expansion, part (b) is direct substitution, and part (c) requires recognizing one linear factor gives x=2 while checking the quadratic's discriminant (b²-4ac = 1-12 < 0) shows no other real roots. All steps are standard textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(p(x) = x^3 + x^2 + 3x - 2x^2 - 2x - 6\) | M1 | Condone one slip |
| \(= x^3 - x^2 + x - 6\) | A1 | \(a = -1\), \(b = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(p(-1) = -3 \times 3\) or \(-1 - 1 - 1 - 6\) | M1 | Must use \(p(-1)\) and not long division |
| (Remainder is) \(-9\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Considering \(x^2 + x + 3 = 0\) and attempting to solve or use discriminant | M1 | \(b^2 - 4ac = 1 - 12 = -11\) |
| \(b^2 - 4ac < 0 \Rightarrow\) no real roots | A1 | CSO |
| Only real root is 2 | B1 | \(x = 2\) |
## Question 6:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $p(x) = x^3 + x^2 + 3x - 2x^2 - 2x - 6$ | M1 | Condone one slip |
| $= x^3 - x^2 + x - 6$ | A1 | $a = -1$, $b = 1$ |
**Part (b)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $p(-1) = -3 \times 3$ or $-1 - 1 - 1 - 6$ | M1 | Must use $p(-1)$ and not long division |
| (Remainder is) $-9$ | A1 | |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Considering $x^2 + x + 3 = 0$ and attempting to solve or use discriminant | M1 | $b^2 - 4ac = 1 - 12 = -11$ |
| $b^2 - 4ac < 0 \Rightarrow$ no real roots | A1 | **CSO** |
| Only real root is 2 | B1 | $x = 2$ |
**Total: 7 marks**
---6 The cubic polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = ( x - 2 ) \left( x ^ { 2 } + x + 3 \right)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { p } ( x )$ can be written in the form $x ^ { 3 } + a x ^ { 2 } + b x - 6$, where $a$ and $b$ are constants whose values are to be found.
\item Use the Remainder Theorem to find the remainder when $\mathrm { p } ( x )$ is divided by $x + 1$.\\
(2 marks)
\item Prove that the equation $( x - 2 ) \left( x ^ { 2 } + x + 3 \right) = 0$ has only one real root and state its value.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2005 Q6 [7]}}