| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with given sum conditions |
| Difficulty | Challenging +1.2 This is a standard Further Maths roots of polynomials question requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) uses the identity (Σα)² = Σα² + 2Σαβ, part (b) applies Σ(1/α) = Σαβ/αβγ, and part (c) expands the product using symmetric functions. While it requires multiple steps and careful algebra, the techniques are well-practiced in FM syllabi with no novel insight needed—slightly above average difficulty due to the multi-part nature and FM content. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(4x^3 + px^2 - 14x + q = 0 \Rightarrow x^3 + \frac{p}{4}x^2 - \frac{14}{4}x + \frac{q}{4} = 0\); \(\alpha+\beta+\gamma = -\frac{p}{4}\), \(\alpha\beta+\alpha\gamma+\beta\gamma = -\frac{14}{4}\) or \(-\frac{7}{2}\) | B1 | Identifies correct values for sum and pair sum |
| \((\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\alpha\gamma+\beta\gamma)\); \(\left(-\frac{p}{4}\right)^2 = 16 + 2\left(-\frac{7}{2}\right) \Rightarrow p = \ldots\) OR \(\left(-\frac{p}{4}\right)^2 - 2\left(-\frac{7}{2}\right) = 16 \Rightarrow p = \ldots\) | M1 | Uses correct identity with their sum and pair sum to find \(p\) |
| \(p = 12\) | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}\) | M1 | Establishes correct identity |
| \(\frac{\left(-\frac{7}{2}\right)}{\left(\frac{-q}{4}\right)} = \frac{14}{3} \Rightarrow q = \ldots\) | M1 | Uses identity with pair sum and product of roots |
| \(q = 3\) | A1 | Allow from incorrect sign of both pair sum and product |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(4\left(\frac{1}{w}\right)^3 + 12\left(\frac{1}{w}\right)^2 - 14\left(\frac{1}{w}\right) + q = 0\) | M1 | Uses \(x = \frac{1}{w}\) substitution |
| \(qw^3 - 14w^2 + 12w + 4 = 0 \Rightarrow \frac{14}{3} = -\frac{-14}{q} \Rightarrow q = \ldots\) | M1 | Simplifies to cubic form \(aw^3+bw^2+cw+d=0\) and uses \(\frac{14}{3} = -\frac{b}{a}\) |
| \(q = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((\alpha-1)(\beta-1)(\gamma-1) = \alpha\beta\gamma - (\alpha\beta+\alpha\gamma+\beta\gamma) + (\alpha+\beta+\gamma) - 1\) | M1, A1 | M1: attempts to multiply out three brackets; A1: correct expansion |
| \(= \left(-\frac{\text{their }3}{4}\right) - \left(-\frac{7}{2}\right) + \left(-\frac{12}{4}\right) - 1 = \ldots\) | dM1 | Dependent on previous M1; substitutes values; condone a slip |
| \(= -\frac{5}{4}\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(4(x+1)^3 + 12(x+1)^2 - 14(x+1) + \text{'3'} = 0\) or substitutes \(x=1\) | M1 | |
| \(= \ldots 4 + \ldots 12 + \ldots -14 + \text{'3'} = 5\) or \(4x^3+24x^2+22x+2+\text{'their }q\text{'}\) | A1ft | Correct constant terms, follow through on their \(q\) |
| \(= -\frac{\text{their constant}}{4}\) | dM1 | |
| \(= -\frac{5}{4}\) | A1 |
## Question 6:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $4x^3 + px^2 - 14x + q = 0 \Rightarrow x^3 + \frac{p}{4}x^2 - \frac{14}{4}x + \frac{q}{4} = 0$; $\alpha+\beta+\gamma = -\frac{p}{4}$, $\alpha\beta+\alpha\gamma+\beta\gamma = -\frac{14}{4}$ or $-\frac{7}{2}$ | B1 | Identifies correct values for sum and pair sum |
| $(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\alpha\gamma+\beta\gamma)$; $\left(-\frac{p}{4}\right)^2 = 16 + 2\left(-\frac{7}{2}\right) \Rightarrow p = \ldots$ OR $\left(-\frac{p}{4}\right)^2 - 2\left(-\frac{7}{2}\right) = 16 \Rightarrow p = \ldots$ | M1 | Uses correct identity with their sum and pair sum to find $p$ |
| $p = 12$ | A1* | cso |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}$ | M1 | Establishes correct identity |
| $\frac{\left(-\frac{7}{2}\right)}{\left(\frac{-q}{4}\right)} = \frac{14}{3} \Rightarrow q = \ldots$ | M1 | Uses identity with pair sum and product of roots |
| $q = 3$ | A1 | Allow from incorrect sign of both pair sum and product |
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $4\left(\frac{1}{w}\right)^3 + 12\left(\frac{1}{w}\right)^2 - 14\left(\frac{1}{w}\right) + q = 0$ | M1 | Uses $x = \frac{1}{w}$ substitution |
| $qw^3 - 14w^2 + 12w + 4 = 0 \Rightarrow \frac{14}{3} = -\frac{-14}{q} \Rightarrow q = \ldots$ | M1 | Simplifies to cubic form $aw^3+bw^2+cw+d=0$ and uses $\frac{14}{3} = -\frac{b}{a}$ |
| $q = 3$ | A1 | |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(\alpha-1)(\beta-1)(\gamma-1) = \alpha\beta\gamma - (\alpha\beta+\alpha\gamma+\beta\gamma) + (\alpha+\beta+\gamma) - 1$ | M1, A1 | M1: attempts to multiply out three brackets; A1: correct expansion |
| $= \left(-\frac{\text{their }3}{4}\right) - \left(-\frac{7}{2}\right) + \left(-\frac{12}{4}\right) - 1 = \ldots$ | dM1 | Dependent on previous M1; substitutes values; condone a slip |
| $= -\frac{5}{4}$ | A1 | Correct value |
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $4(x+1)^3 + 12(x+1)^2 - 14(x+1) + \text{'3'} = 0$ or substitutes $x=1$ | M1 | |
| $= \ldots 4 + \ldots 12 + \ldots -14 + \text{'3'} = 5$ or $4x^3+24x^2+22x+2+\text{'their }q\text{'}$ | A1ft | Correct constant terms, follow through on their $q$ |
| $= -\frac{\text{their constant}}{4}$ | dM1 | |
| $= -\frac{5}{4}$ | A1 | |
---\begin{enumerate}
\item The cubic equation
\end{enumerate}
$$4 x ^ { 3 } + p x ^ { 2 } - 14 x + q = 0$$
where $p$ and $q$ are real positive constants, has roots $\alpha , \beta$ and $\gamma$\\
Given that $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = 16$\\
(a) show that $p = 12$
Given that $\frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma } = \frac { 14 } { 3 }$\\
(b) determine the value of $q$
Without solving the cubic equation,\\
(c) determine the value of $( \alpha - 1 ) ( \beta - 1 ) ( \gamma - 1 )$
\hfill \mbox{\textit{Edexcel CP2 2022 Q6 [10]}}