FP1 June 2015 Q1
1. \[\mathrm{f}(x) = 9x^3 - 33x^2 - 55x - 25\]
Given that \(x = 5\) is a solution of the equation \(\mathrm{f}(x) = 0\), use an algebraic method to solve \(\mathrm{f}(x) = 0\) completely. (5)
| Scheme | Marks |
|---|---|
| \((x - 5)\) is a factor of \(\mathrm{f}(x)\) so \(\mathrm{f}(x) = (x - 5)(9x^2 \ldots\ )\) | M1 |
| \(\mathrm{f}(x) = (x - 5)(9x^2 + 12x + 5)\) | A1 |
| Solve \((9x^2 + 12x + 5) = 0\) to give \(x =\) | M1 |
| \((x =) -\dfrac{2}{3} - \dfrac{1}{3}\mathrm{i}, \ -\dfrac{2}{3} + \dfrac{1}{3}\mathrm{i}\) or \(-\dfrac{2}{3} \pm \dfrac{1}{3}\mathrm{i}\) or \(\dfrac{-2 \pm \mathrm{i}}{3}\) oe (and 5) | A1cao A1ft |
| (5) | |
| (5 marks) |
Notes
M1: Uses \((x - 5)\) as factor and begins division or process to obtain quadratic with \(9x^2\). Award if no working but quadratic factor completely correct.
A1: \(9x^2 + 12x + 5\)
M1: Solves their quadratic by usual rules leading to \(x =\)
Award if one complex root correct with no working.
Award for \((9x^2 + \ldots\) incorrectly factorised to \((3x + p)(3x + q)\), where \(|pq| = 5\)
A1: One correct complex root. Accept any exact equivalent form. Accept single fraction and \(\pm\)
A1ft: Conjugate of their first complex root.