| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question on polynomial roots requiring substitution, symmetric functions, and recurrence relations. While it involves several steps and the power sums technique, the substitution is guided, and the methods are standard for FP1. The conceptual demand is moderate—harder than typical A-level but routine for Further Maths students who have practiced these techniques. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = (y+1)^3\) | M1 | Obtains an equation in \(y\) not involving radicals |
| \(y = y^3 + 3y^2 + 3y + 1 \Rightarrow y^3 + 3y^2 + 2y + 1 = 0\) | A1 | AG |
| \(S_3 = -3\) | B1 |
| Answer | Marks |
|---|---|
| \(S_{-3} = \frac{\alpha^3\beta^3 + \beta^3\gamma^3 + \alpha^3\gamma^3}{\alpha^3\beta^3\gamma^3} = \frac{2}{-1} = -2\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_6 = (-3)^2 - 2(2) = 5\) | M1 A1 | Uses \((\sum\alpha)^2 = \sum\alpha^2 + 2\sum_{\alpha\neq\beta}\alpha\beta\). AG |
| \(S_9 = -3S_6 - 2S_3 - 3 = -3(5) - 2(-3) - 3 = -12\) | M1 A1 | Sums \(y^3 - 3y^2 + 2y - 1 = 0\) for \(y = \alpha^3, \beta^3, \gamma^3\) |
## Question 6(i):
| $y = (y+1)^3$ | M1 | Obtains an equation in $y$ not involving radicals |
|---|---|---|
| $y = y^3 + 3y^2 + 3y + 1 \Rightarrow y^3 + 3y^2 + 2y + 1 = 0$ | A1 | AG |
| $S_3 = -3$ | B1 | |
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## Question 6(ii):
| $S_{-3} = \frac{\alpha^3\beta^3 + \beta^3\gamma^3 + \alpha^3\gamma^3}{\alpha^3\beta^3\gamma^3} = \frac{2}{-1} = -2$ | M1 A1 | |
|---|---|---|
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## Question 6(iii):
| $S_6 = (-3)^2 - 2(2) = 5$ | M1 A1 | Uses $(\sum\alpha)^2 = \sum\alpha^2 + 2\sum_{\alpha\neq\beta}\alpha\beta$. AG |
|---|---|---|
| $S_9 = -3S_6 - 2S_3 - 3 = -3(5) - 2(-3) - 3 = -12$ | M1 A1 | Sums $y^3 - 3y^2 + 2y - 1 = 0$ for $y = \alpha^3, \beta^3, \gamma^3$ |
---6 The equation
$$x ^ { 3 } - x + 1 = 0$$
has roots $\alpha , \beta , \gamma$.\\
(i) Use the relation $x = y ^ { \frac { 1 } { 3 } }$ to show that the equation
$$y ^ { 3 } + 3 y ^ { 2 } + 2 y + 1 = 0$$
has roots $\alpha ^ { 3 } , \beta ^ { 3 } , \gamma ^ { 3 }$. Hence write down the value of $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }$.\\
Let $S _ { n } = \alpha ^ { n } + \beta ^ { n } + \gamma ^ { n }$.\\
(ii) Find the value of $S _ { - 3 }$.\\
(iii) Show that $S _ { 6 } = 5$ and find the value of $S _ { 9 }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q6 [9]}}