| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with special relationships |
| Difficulty | Standard +0.3 This is a straightforward application of Vieta's formulas with a helpful constraint. The product of roots gives a³=216 immediately (so a=6), then the sum gives 21a=21 confirming a=6, and finally the sum of products yields k. The geometric progression structure makes this easier than a generic roots problem, requiring only basic algebraic manipulation rather than problem-solving insight. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha\beta\gamma = a^3 = 216 \Rightarrow a = 6\) | M1 A1 | Uses product of roots |
| \(a + ar + ar^{-1} = 21\), \(6(1 + r + r^{-1}) = 21\) | M1 | Uses sum of roots |
| \(2r^2 - 5r + 2 = 0 \Rightarrow r = 2\) or \(r = 0.5\) | M1 A1 | Substitutes for \(a\) and solves quadratic |
| Roots are 6, 12, 3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k = \alpha\beta + \alpha\gamma + \beta\gamma = 6(12) + 6(3) + 12(3) = 126\) | M1 A1 | Or finds coefficient of \(x\) in \((x-3)(x-6)(x-12)\). Or substitutes root into equation |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha\beta\gamma = a^3 = 216 \Rightarrow a = 6$ | M1 A1 | Uses product of roots |
| $a + ar + ar^{-1} = 21$, $6(1 + r + r^{-1}) = 21$ | M1 | Uses sum of roots |
| $2r^2 - 5r + 2 = 0 \Rightarrow r = 2$ or $r = 0.5$ | M1 A1 | Substitutes for $a$ and solves quadratic |
| Roots are 6, 12, 3 | A1 | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = \alpha\beta + \alpha\gamma + \beta\gamma = 6(12) + 6(3) + 12(3) = 126$ | M1 A1 | Or finds coefficient of $x$ in $(x-3)(x-6)(x-12)$. Or substitutes root into equation |4 It is given that the equation
$$x ^ { 3 } - 21 x ^ { 2 } + k x - 216 = 0$$
where $k$ is a constant, has real roots $a , a r$ and $a r ^ { - 1 }$.\\
(i) Find the numerical values of the roots.\\
(ii) Deduce the value of $k$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q4 [8]}}