AS June 2018 Paper 1 Q4
4 Find a cubic equation with real coefficients, two of whose roots are \(2 - \mathrm{i}\) and 3. [5]
| Scheme | Marks | AO |
|---|---|---|
| third root is \(2 + \mathrm{i}\) | B1 | 1.2 |
| \(2 - \mathrm{i} + 2 + \mathrm{i} + 3 = 7\) | B1ft | 3.1a |
| \((2 - \mathrm{i})(2 + \mathrm{i}) + (2 + \mathrm{i})3 + 3(2 - \mathrm{i}) = 17\) | B1ft | 1.1 |
| \((2 - \mathrm{i})(2 + \mathrm{i})3 = 15\) | B1ft | 1.1 |
| so eqn is \(z^3 - 7z^2 + 17z - 15 = 0\) | B1cao | 1.1 |
| [5] |
Notes
B1: soi
B1ft: (1st) ft their \(2 + \mathrm{i}\)
B1cao: must include ‘\(= 0\)’
any variable, e.g. \(x\)
Alternative
| Scheme | Marks |
|---|---|
| \((z - 2 + \mathrm{i})(z - 2 - \mathrm{i}) = z^2 - 4z + 5\) | M1 A1 |
| \((z^2 - 4z + 5)(z - 3) = z^3 - 7z^2 + 17z - 15\) so eqn is \(z^3 - 7z^2 + 17z - 15 = 0\) | M1 A1 |
M1: (1st) \((z - 2 + \mathrm{i})(z - 2 - \mathrm{i})\)
A1: (1st) \(= z^2 - 4z + 5\)
M1: (2nd) their \((z^2 - 4z + 5) \times (z - 3)\)