Standard +0.3 This is a straightforward application of the complex conjugate root theorem followed by polynomial division. Given one complex root with real coefficients, students immediately know the conjugate is also a root, can form a quadratic factor, then divide to find the remaining real root. This is a standard FP1 exercise requiring routine algebraic manipulation rather than problem-solving insight.
2.
$$f ( z ) = z ^ { 3 } + 5 z ^ { 2 } + 11 z + 15$$
Given that \(z = 2 i - 1\) is a solution of the equation \(f ( z ) = 0\), use algebra to solve \(f ( z ) = 0\) completely.
(5)
First M for expanding their \((z-\alpha)(z-\beta)\)
\((z+3)(z^2+2z+5)=0\)
M1
Second M for inspection or long division
\(z=-3\)
A1
(5) [5]
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = -2i-1$ is also a root | B1 | |
| $(z-(2i-1))(z-(-2i-1)) = z^2+2z+5$ | M1A1 | First M for expanding their $(z-\alpha)(z-\beta)$ |
| $(z+3)(z^2+2z+5)=0$ | M1 | Second M for inspection or long division |
| $z=-3$ | A1 | |
| | **(5) [5]** | |
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2.
$$f ( z ) = z ^ { 3 } + 5 z ^ { 2 } + 11 z + 15$$
Given that $z = 2 i - 1$ is a solution of the equation $f ( z ) = 0$, use algebra to solve $f ( z ) = 0$ completely.\\
(5)\\
\hfill \mbox{\textit{Edexcel FP1 2013 Q2 [5]}}