A2 June 2024 Paper 2 Q6
6 The cubic equation
\[x^3 + 5x^2 - 4x + 2 = 0\]has roots \(\alpha\), \(\beta\) and \(\gamma\)
Find a cubic equation, with integer coefficients, whose roots are \(3\alpha\), \(3\beta\) and \(3\gamma\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Sets \(y = 3x\) PI by correct substitution or Obtains one of \(\sum 3\alpha = \pm 15\) \(\sum (3\alpha)(3\beta) = \pm 36\) \((3\alpha)(3\beta)(3\gamma) = \pm 54\) | M1 | 1.1a |
| Replaces \(x\) with \(\dfrac{y}{3}\) or \(3y\) Accept \(x\) for \(y\) or Obtains at least two of \(\sum 3\alpha = \pm 15\) \(\sum (3\alpha)(3\beta) = \pm 36\) \((3\alpha)(3\beta)(3\gamma) = \pm 54\) | M1 | 1.1a |
| Obtains \(y^3 + 15y^2 - 36y + 54 = 0\) OE with integer coefficients. | A1 | 1.1b |
| (3 marks) |
Typical solution
Let \(y = 3x\)
Then \(x = \dfrac{y}{3}\)
\[\frac{y^3}{27} + \frac{5y^2}{9} - \frac{4y}{3} + 2 = 0\]\[y^3 + 15y^2 - 36y + 54 = 0\]