| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Standard +0.3 This is a standard textbook exercise on roots of polynomials requiring application of Vieta's formulas and a routine transformation. Part (a) uses the formula 1/α + 1/β + 1/γ = -coefficient of x/constant term. Part (b) involves the standard technique of substituting y = 1/x to find the transformed equation. Both parts are mechanical applications of well-practiced techniques with no novel insight required, making this slightly easier than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha\beta\gamma = -\frac{1}{3}\) and \(\alpha\beta+\alpha\gamma+\beta\gamma = -\frac{4}{3}\) | B1 | Correct values for product and pair sum of roots |
| \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} = \frac{-\frac{4}{3}}{-\frac{1}{3}}\) | M1 | Complete method; must substitute values of product and pair sum |
| \(= 4\) | A1 | Correct value 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| New product \(= \frac{1}{\alpha}\times\frac{1}{\beta}\times\frac{1}{\gamma} = \frac{1}{\alpha\beta\gamma} = \frac{1}{-\frac{1}{3}} = \ldots(-3)\); New pair sum \(\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\alpha\gamma} = \frac{\gamma+\alpha+\beta}{\alpha\beta\gamma} = \frac{-\frac{1}{3}}{-\frac{1}{3}} = \ldots(1)\) | M1 | Correct method to find new pair sum and new product |
| \(x^3 - (\text{part (a)})x^2 + (\text{new pair sum})x - (\text{new product}) = 0\) | M1 | Applies correct cubic structure using their values |
| \(x^3 - 4x^2 + x + 3 = 0\) | A1 | Fully correct equation including \(=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(z = \frac{1}{x} \Rightarrow \frac{3}{x^3}+\frac{1}{x^2}-\frac{4}{x}+1=0\) | M1 | Realises connection between roots and substitutes into cubic |
| \(x^3 - 4x^2 + x + 3 = 0\) | M1, A1 | Manipulates into form \(x^3+ax^2+bx+c=0\); fully correct equation including \(=0\) |
## Question 9:
**Part (a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha\beta\gamma = -\frac{1}{3}$ and $\alpha\beta+\alpha\gamma+\beta\gamma = -\frac{4}{3}$ | B1 | Correct values for product and pair sum of roots |
| $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} = \frac{-\frac{4}{3}}{-\frac{1}{3}}$ | M1 | Complete method; must substitute values of product and pair sum |
| $= 4$ | A1 | Correct value 4 |
Note: If candidate does not divide by 3 so that $\alpha\beta\gamma=-1$ and $\alpha\beta+\alpha\gamma+\beta\gamma=-4$, maximum score is B0 M1 A0.
**Part (b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| New product $= \frac{1}{\alpha}\times\frac{1}{\beta}\times\frac{1}{\gamma} = \frac{1}{\alpha\beta\gamma} = \frac{1}{-\frac{1}{3}} = \ldots(-3)$; New pair sum $\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\alpha\gamma} = \frac{\gamma+\alpha+\beta}{\alpha\beta\gamma} = \frac{-\frac{1}{3}}{-\frac{1}{3}} = \ldots(1)$ | M1 | Correct method to find new pair sum and new product |
| $x^3 - (\text{part (a)})x^2 + (\text{new pair sum})x - (\text{new product}) = 0$ | M1 | Applies correct cubic structure using their values |
| $x^3 - 4x^2 + x + 3 = 0$ | A1 | Fully correct equation including $=0$ |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $z = \frac{1}{x} \Rightarrow \frac{3}{x^3}+\frac{1}{x^2}-\frac{4}{x}+1=0$ | M1 | Realises connection between roots and substitutes into cubic |
| $x^3 - 4x^2 + x + 3 = 0$ | M1, A1 | Manipulates into form $x^3+ax^2+bx+c=0$; fully correct equation including $=0$ |
---\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$3 x ^ { 3 } + x ^ { 2 } - 4 x + 1 = 0$$
has roots $\alpha , \beta$, and $\gamma$.\\
Without solving the cubic equation,\\
(a) determine the value of $\frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma }$\\
(b) find a cubic equation that has roots $\frac { 1 } { \alpha } , \frac { 1 } { \beta }$ and $\frac { 1 } { \gamma }$, giving your answer in the form $x ^ { 3 } + a x ^ { 2 } + b x + c = 0$, where $a , b$ and $c$ are integers to be determined.
\hfill \mbox{\textit{Edexcel CP AS 2020 Q9 [6]}}