AS October 2021 Paper 1 Q3
3 In this question you must show detailed reasoning.
The equation \(x^4 - 7x^3 - 2x^2 + 218x - 1428 = 0\) has a root \(3 - 5\mathrm{i}\).
Find the other three roots of this equation. [6]
| Scheme | Marks | AO |
|---|---|---|
| \(3 + 5\mathrm{i}\) is a root | B1 | 1.2 |
| Attempt to expand \((x - (3 + 5\mathrm{i}))(x - (3 - 5\mathrm{i}))\) | M1 | 1.1 |
| \(= x^2 - 6x + 34\) so this must be a factor | A1 | 2.2a |
| \(x^4 - 7x^3 - 2x^2 + 218x - 1428 =\) \((x^2 - 6x + 34)(x^2 + \ldots x - 42)\) or \((x^2 - 6x + 34)(x^2 - x + \ldots)\) | M1 | 1.1 |
| \((x^2 - 6x + 34)(x^2 - x - 42)\) | A1 | 1.1 |
| \((x^2 - x - 42) = (x - 7)(x + 6) \Rightarrow\) roots \(-6\), \(7\) (and \(3 + 5\mathrm{i}\)) | A1 | 1.1 |
| [6] |
Notes
B1: Need to see statement that \(3 + 5\mathrm{i}\) is a root.
May happen at end of question
M1: (1st) Attempt to use the conjugate pair to derive a real quadratic
May see \((3 + 5\mathrm{i})(3 - 5\mathrm{i}) = 9 + 25 = 34\) and \((3 + 5\mathrm{i}) + (3 - 5\mathrm{i}) = 6\) instead of expansion
M1: (2nd) Attempt to factorise or divide resulting in \(x^2\) and one other term
NB: This question required detailed reasoning
A1: (3rd) \(3 + 5\mathrm{i}\) may be mentioned as a root earlier in the solution