| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Symmetric functions of roots |
| Difficulty | Standard +0.8 This is a Core Pure AS question on symmetric functions of roots requiring multiple techniques: using Vieta's formulas, the identity (Σα)² = Σα² + 2Σαβ for part (i), recognizing Σ(1/α) = -coefficient manipulation for part (ii), and polynomial substitution P(3) for part (iii). While the techniques are standard for Further Maths students, the multi-part nature and need to apply three different approaches makes it moderately challenging, placing it above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sum \alpha_i = -\frac{5}{3}\) and \(\sum \alpha_i\alpha_j = 0\) | B1 | Correct sum and pair sum of roots seen or implied. Must realise pair sum is zero. Can be awarded if seen in parts (ii) or (iii) |
| \(\alpha^2+\beta^2+\gamma^2+\delta^2 = (\alpha+\beta+\gamma+\delta)^2 - 2\left(\sum \alpha_i\alpha_j\right)\) | M1 | Uses correct expression for sum of squares |
| \(= \frac{25}{9} - 2\times0 = \frac{25}{9}\) | A1 | Allow from incorrect sign on sum of squares (but B0 if sign incorrect) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sum \alpha_i\alpha_j\alpha_k = \frac{7}{3}\) and \(\prod \alpha_i = 2\), or \(x = \frac{2}{w}\) used in equation | B1 | Correct triple sum and product of roots. Can be awarded if seen in parts (i) or (iii) |
| \(2\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}\right) = 2\times\frac{\sum \alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = 2\times\frac{\frac{7}{3}}{\frac{6}{3}}\) | M1 | Substitutes values into expression. In alternative, rearranges equation to quartic in \(w\) and finds sum of roots |
| \(= 2\times\frac{7/3}{2} = \frac{14}{6} = \frac{7}{3}\) | A1 | Allow from incorrect sign of both triple sum and product (but B0 if sign incorrect) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((3-\alpha)(3-\beta)(3-\gamma)(3-\delta) = \ldots\) expands all four brackets, or equation with roots \(3(3-x)^4+5(3-x)^3-7(3-x)+6=0\) | M1 | Correct method to find the value — may recognise structure, expand in stages, or use linear transformation \((3-x)\) |
| \(= 81 - 27\left(\sum\alpha_i\right) + 9\left(\sum\alpha_i\alpha_j\right) - 3\left(\sum\alpha_i\alpha_j\alpha_k\right) + \prod\alpha_i\) \(= 81 - 27\left(-\frac{5}{3}\right) + 9(0) - 3\left(\frac{7}{3}\right) + 2\) | dM1 | Dependent on M1. Uses at least 2 values of sum of roots etc. in expression |
| \(= 121\) | A1 |
# Question 4:
## Part (i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sum \alpha_i = -\frac{5}{3}$ and $\sum \alpha_i\alpha_j = 0$ | B1 | Correct sum and pair sum of roots seen or implied. Must realise pair sum is zero. Can be awarded if seen in parts (ii) or (iii) |
| $\alpha^2+\beta^2+\gamma^2+\delta^2 = (\alpha+\beta+\gamma+\delta)^2 - 2\left(\sum \alpha_i\alpha_j\right)$ | M1 | Uses correct expression for sum of squares |
| $= \frac{25}{9} - 2\times0 = \frac{25}{9}$ | A1 | Allow from incorrect sign on sum of squares (but B0 if sign incorrect) |
**(3 marks)**
## Part (ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sum \alpha_i\alpha_j\alpha_k = \frac{7}{3}$ and $\prod \alpha_i = 2$, or $x = \frac{2}{w}$ used in equation | B1 | Correct triple sum and product of roots. Can be awarded if seen in parts (i) or (iii) |
| $2\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}\right) = 2\times\frac{\sum \alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = 2\times\frac{\frac{7}{3}}{\frac{6}{3}}$ | M1 | Substitutes values into expression. In alternative, rearranges equation to quartic in $w$ and finds sum of roots |
| $= 2\times\frac{7/3}{2} = \frac{14}{6} = \frac{7}{3}$ | A1 | Allow from incorrect sign of both triple sum and product (but B0 if sign incorrect) |
**(3 marks)**
## Part (iii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(3-\alpha)(3-\beta)(3-\gamma)(3-\delta) = \ldots$ expands all four brackets, or equation with roots $3(3-x)^4+5(3-x)^3-7(3-x)+6=0$ | M1 | Correct method to find the value — may recognise structure, expand in stages, or use linear transformation $(3-x)$ |
| $= 81 - 27\left(\sum\alpha_i\right) + 9\left(\sum\alpha_i\alpha_j\right) - 3\left(\sum\alpha_i\alpha_j\alpha_k\right) + \prod\alpha_i$ $= 81 - 27\left(-\frac{5}{3}\right) + 9(0) - 3\left(\frac{7}{3}\right) + 2$ | dM1 | Dependent on M1. Uses at least 2 values of sum of roots etc. in expression |
| $= 121$ | A1 | |
**(3 marks)**
---\begin{enumerate}
\item The roots of the quartic equation
\end{enumerate}
$$3 x ^ { 4 } + 5 x ^ { 3 } - 7 x + 6 = 0$$
are $\alpha , \beta , \gamma$ and $\delta$\\
Making your method clear and without solving the equation, determine the exact value of\\
(i) $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 }$\\
(ii) $\frac { 2 } { \alpha } + \frac { 2 } { \beta } + \frac { 2 } { \gamma } + \frac { 2 } { \delta }$\\
(iii) $( 3 - \alpha ) ( 3 - \beta ) ( 3 - \gamma ) ( 3 - \delta )$
\hfill \mbox{\textit{Edexcel CP AS 2022 Q4 [9]}}