FP1 June 2018 Q1
1. \[\mathrm{f}(z) = 2z^3 - 4z^2 + 15z - 13\]
Given that \(\mathrm{f}(z) \equiv (z - 1)(2z^2 + az + b)\), where \(a\) and \(b\) are real constants,
(a) find the value of \(a\) and the value of \(b\). (2)
(b) Hence use algebra to find the three roots of the equation \(\mathrm{f}(z) = 0\) (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(z) = 2z^3 - 4z^2 + 15z - 13 \equiv (z - 1)(2z^2 + az + b)\) | |
| \(a = -2,\ b = 13\) | B1 B1 |
| (2) |
Notes
B1: At least one of either \(a = -2\) or \(b = 13\) or seen as their coefficients.
B1: Both \(a = -2\) and \(b = 13\) or seen as their coefficients.
| Scheme | Marks |
|---|---|
| \(\{z =\}\) 1 is a root | B1 |
| \(\left\{2z^2 - 2z + 13 = 0 \Rightarrow z^2 - z + \dfrac{13}{2} = 0\right\}\) | |
| Either • \(z = \dfrac{2 \pm \sqrt{4 - 4(2)(13)}}{2(2)}\) or • \(\left(z - \dfrac{1}{2}\right)^2 - \dfrac{1}{4} + \dfrac{13}{2} = 0\) and \(z = \ldots\) or • \((2z - 1)^2 - 1 + 26 = 0\) and \(z = \ldots\) | M1 |
| So, \(\{z =\}\ \dfrac{1}{2} + \dfrac{5}{2}\mathrm{i},\ \dfrac{1}{2} - \dfrac{5}{2}\mathrm{i}\) | A1 A1ft |
| (4) | |
| (6 marks) |
Notes
B1: 1 is a root, seen anywhere.
M1: Correct method for solving a 3-term quadratic equation. Do not allow M1 here for an attempt at factorising.
A1: At least one of either \(\dfrac{1}{2} + \dfrac{5}{2}\mathrm{i}\) or \(\dfrac{1}{2} - \dfrac{5}{2}\mathrm{i}\) or any equivalent form.
A1ft: For conjugate of first complex root
(corrected from the printed mark scheme: the third method is printed as \((2z - 1)^2 - 1 + +13 = 0\))