| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Standard +0.3 This is a standard Further Maths question on transformed roots requiring routine application of Vieta's formulas and symmetric function identities. While it involves multiple steps (finding sum/product, expanding, using α³+β³ identity, forming new equation), each step follows a well-established procedure taught in FP1 with no novel problem-solving required. Slightly easier than average A-level difficulty due to its formulaic nature. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\alpha + \beta = 2\), \(\alpha\beta = \frac{2}{3}\) | B1B1 (2) | SC 1/2 for answers 6 and 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((\alpha+\beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3\) | B1 (1) | Accept unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\) | M1 | |
| Substitution of numerical values | m1 | |
| \(\alpha^3 + \beta^3 = 4\) | A1 (3) | Convincingly shown AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\alpha^3\beta^3 = \frac{8}{27}\) | B1 | |
| Equation of form \(px^2 \pm 4px + r = 0\) | M1 | |
| \(27x^2 - 108x + 8 = 0\) | A1\(\checkmark\) (3) | ft wrong value for \(\alpha^3\beta^3\) |
## Question 1:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\alpha + \beta = 2$, $\alpha\beta = \frac{2}{3}$ | B1B1 (2) | SC 1/2 for answers 6 and 2 |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $(\alpha+\beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3$ | B1 (1) | Accept unsimplified |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$ | M1 | |
| Substitution of numerical values | m1 | |
| $\alpha^3 + \beta^3 = 4$ | A1 (3) | Convincingly shown AG |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\alpha^3\beta^3 = \frac{8}{27}$ | B1 | |
| Equation of form $px^2 \pm 4px + r = 0$ | M1 | |
| $27x^2 - 108x + 8 = 0$ | A1$\checkmark$ (3) | ft wrong value for $\alpha^3\beta^3$ |
---1 The quadratic equation
$$3 x ^ { 2 } - 6 x + 2 = 0$$
has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the numerical values of $\alpha + \beta$ and $\alpha \beta$.
\item \begin{enumerate}[label=(\roman*)]
\item Expand $( \alpha + \beta ) ^ { 3 }$.
\item Show that $\alpha ^ { 3 } + \beta ^ { 3 } = 4$.
\end{enumerate}\item Find a quadratic equation with roots $\alpha ^ { 3 }$ and $\beta ^ { 3 }$, giving your answer in the form $p x ^ { 2 } + q x + r = 0$, where $p , q$ and $r$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2006 Q1 [9]}}