| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Symmetric functions of roots |
| Difficulty | Standard +0.3 This is a standard symmetric functions question requiring routine application of Vieta's formulas. Part (i) uses the identity α²+β²+γ² = (α+β+γ)²-2(αβ+βγ+γα), part (ii) simplifies to 3(α+β+γ)/αβγ, and part (iii) involves substituting x=5 into the polynomial. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha + \beta + \gamma = \frac{3}{2}\), \(\alpha\beta + \alpha\gamma + \beta\gamma = \frac{5}{2}\) | B1 | 3.1a |
| \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\alpha\gamma+\beta\gamma) = \left(\frac{3}{2}\right)^2 - 2\left(\frac{5}{2}\right) = \ldots\) | M1 | 1.1b |
| \(= -\frac{11}{4} = -2.75\) | A1 cso | 1.1b |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha\beta\gamma = -\frac{7}{2}\) or \(x = \frac{3}{w}\) used in equation | B1 | 2.2a |
| \(\frac{3}{\alpha}+\frac{3}{\beta}+\frac{3}{\gamma} = \frac{3(\alpha\beta+\alpha\gamma+\beta\gamma)}{\alpha\beta\gamma} = \frac{3\left(\frac{5}{2}\right)}{\left(-\frac{7}{2}\right)}\) or \(2\left(\frac{3}{w}\right)^3 - 3\left(\frac{3}{w}\right)^2 + 5\left(\frac{3}{w}\right) + 7 = 0 \Rightarrow 7w^3+15w^2-27w+54\{=0\} \Rightarrow -\frac{`15`}{`7`}\) | M1 | 1.1b |
| \(= -\frac{15}{7}\) cso | A1 | 1.1b |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((5-\alpha)(5-\beta)(5-\gamma) = A \pm B(\alpha+\beta+\gamma) \pm C(\alpha\beta+\alpha\gamma+\beta\gamma) \pm (\alpha\beta\gamma)\) \(= \left\{5^3 - 5^2(\alpha+\beta+\gamma) + 5(\alpha\beta+\alpha\gamma+\beta\gamma) - \alpha\beta\gamma\right\}\) or \(2(5-w)^3-3(5-w)^2+5(5-w)+7\{=0\}\) or \(f(x)=A(x-\alpha)(x-\beta)(x-\gamma) \Rightarrow A=2\) | M1 | 3.1a |
| \((5-\alpha)(5-\beta)(5-\gamma) = 125 - 25\left(\frac{3}{2}\right)+5\left(\frac{5}{2}\right)+\frac{7}{2}\) or \(= -\left(\frac{2\times125-3\times25+25+7}{-2}\right)\) or \(-2w^3+27w^2-125w+207\{=0\} \Rightarrow -\frac{`207`}{`-2`}\) or \(f(5)=2(5-\alpha)(5-\beta)(5-\gamma) \Rightarrow (5-\alpha)(5-\beta)(5-\gamma)=\frac{f(5)}{2}\) | M1 | 1.1b |
## Question 1:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha + \beta + \gamma = \frac{3}{2}$, $\alpha\beta + \alpha\gamma + \beta\gamma = \frac{5}{2}$ | B1 | 3.1a |
| $\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\alpha\gamma+\beta\gamma) = \left(\frac{3}{2}\right)^2 - 2\left(\frac{5}{2}\right) = \ldots$ | M1 | 1.1b |
| $= -\frac{11}{4} = -2.75$ | A1 cso | 1.1b |
| **Total: 3 marks** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha\beta\gamma = -\frac{7}{2}$ or $x = \frac{3}{w}$ used in equation | B1 | 2.2a |
| $\frac{3}{\alpha}+\frac{3}{\beta}+\frac{3}{\gamma} = \frac{3(\alpha\beta+\alpha\gamma+\beta\gamma)}{\alpha\beta\gamma} = \frac{3\left(\frac{5}{2}\right)}{\left(-\frac{7}{2}\right)}$ **or** $2\left(\frac{3}{w}\right)^3 - 3\left(\frac{3}{w}\right)^2 + 5\left(\frac{3}{w}\right) + 7 = 0 \Rightarrow 7w^3+15w^2-27w+54\{=0\} \Rightarrow -\frac{`15`}{`7`}$ | M1 | 1.1b |
| $= -\frac{15}{7}$ cso | A1 | 1.1b |
| **Total: 3 marks** | | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(5-\alpha)(5-\beta)(5-\gamma) = A \pm B(\alpha+\beta+\gamma) \pm C(\alpha\beta+\alpha\gamma+\beta\gamma) \pm (\alpha\beta\gamma)$ $= \left\{5^3 - 5^2(\alpha+\beta+\gamma) + 5(\alpha\beta+\alpha\gamma+\beta\gamma) - \alpha\beta\gamma\right\}$ **or** $2(5-w)^3-3(5-w)^2+5(5-w)+7\{=0\}$ **or** $f(x)=A(x-\alpha)(x-\beta)(x-\gamma) \Rightarrow A=2$ | M1 | 3.1a |
| $(5-\alpha)(5-\beta)(5-\gamma) = 125 - 25\left(\frac{3}{2}\right)+5\left(\frac{5}{2}\right)+\frac{7}{2}$ **or** $= -\left(\frac{2\times125-3\times25+25+7}{-2}\right)$ **or** $-2w^3+27w^2-125w+207\{=0\} \Rightarrow -\frac{`207`}{`-2`}$ **or** $f(5)=2(5-\alpha)(5-\beta)(5-\gamma) \Rightarrow (5-\alpha)(5-\beta)(5-\gamma)=\frac{f(5)}{2}$ | M1 | 1.1b |\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$2 x ^ { 3 } - 3 x ^ { 2 } + 5 x + 7 = 0$$
has roots $\alpha , \beta$ and $\gamma$.\\
Without solving the equation, determine the exact value of\\
(i) $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$\\
(ii) $\frac { 3 } { \alpha } + \frac { 3 } { \beta } + \frac { 3 } { \gamma }$\\
(iii) $( 5 - \alpha ) ( 5 - \beta ) ( 5 - \gamma )$
\hfill \mbox{\textit{Edexcel CP AS 2024 Q1 [9]}}