A2 June 2019 Paper 1 Q3
3 In this question you must show detailed reasoning.
You are given that \(x = 2 + 5\mathrm{i}\) is a root of the equation \(x^3 - 2x^2 + 21x + 58 = 0\).
Solve the equation. [4]
| Scheme | Marks | AO |
|---|---|---|
| DR Second root is the conjugate of \(2 + 5\mathrm{i}\), so \(x = 2 - 5\mathrm{i}\) soi | B1 | 2.2a |
| So the cubic \(x^3 - 2x^2 + 21x + 58 = 0\) can be written \((x - a)(x - (2 + 5\mathrm{i}))(x - (2 - 5\mathrm{i})) = 0\) | M1 | 1.1 |
| \(\Rightarrow -a(2 + 5\mathrm{i})(2 - 5\mathrm{i}) = 58\) \(\Rightarrow a = -2\) | A1 | 2.1 |
| So the solution of the cubic is \(\Rightarrow x = -2,\ 2 \pm 5\mathrm{i}\) | A1 | 1.1 |
| [4] |
Notes
B1: Soi by correct quadratic \(x^2 - 4x + 29\)
M1: Attempt to factorise using complex conjugate.
Any valid method to find real root by reasoning, including division, or listing or using the factor theorem or sum of roots.
A1: Shown convincingly oe (i.e. \((x + 2)\) seen)
NB. A DR question so full reasoning must be shown.