A2 June 2025 Paper 1 Q13
13 The function \(\mathrm{f}\) is defined by
\[\mathrm{f}(z) = 4z^3 + rz^2 + 92z + s\]where \(r\) and \(s\) are real numbers.
One of the roots of the equation \(\mathrm{f}(z) = 0\) is \(-4 + 3\mathrm{i}\)
(a) Find the other two roots of the equation \(\mathrm{f}(z) = 0\) [4 marks]
(b) Find the value of \(r\) and the value of \(s\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(-4 - 3\mathrm{i}\) | B1 | 1.2 |
| Selects a correct method; for example, uses the pairwise sum of roots, or substitutes the given root into the equation, or expands \((z - (-4 + 3\mathrm{i}))(z - (-4 - 3\mathrm{i}))\) | M1 | 3.1a |
| Sets the pairwise sum \(= \pm\dfrac{92}{4}\) or Finds \(\mathrm{f}(z)\) as a product of a quadratic and a linear factor or in expanded form PI by \((4z - 1)\) PI by correct values for \(r\) and \(s\) | M1 | 1.1a |
| Obtains \(\dfrac{1}{4}\) | A1 | 2.2a |
| (4) |
Typical solution
One root is \(-4 - 3\mathrm{i}\)
Let the other root \(= \gamma\)
Then
\[(-4 + 3\mathrm{i})(-4 - 3\mathrm{i}) + \gamma(-4 + 3\mathrm{i}) + \gamma(-4 - 3\mathrm{i}) = \frac{92}{4} = 23\]\[25 - 8\gamma = 23\]\[\gamma = \frac{1}{4}\]So the other roots are \(-4 - 3\mathrm{i}\) and \(\dfrac{1}{4}\)
| Scheme | Marks | AO |
|---|---|---|
| Forms an equation in \(r\), eg uses the sum of roots. or Expands their \((z^2 + 8z + 25)(4z - 1)\) | M1 | 3.1a |
| Forms an equation in \(s\), eg uses the product of roots. or Compares coefficients | M1 | 3.1a |
| Obtains \(r = 31\) and \(s = -25\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |