| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Complex roots with real coefficients |
| Difficulty | Moderate -0.8 This is a standard FP1 question testing the fundamental property that complex roots of polynomials with real coefficients occur in conjugate pairs. Part (a) requires only recall of this fact, and part (b) involves straightforward application of sum and product of roots formulas or expansion. No problem-solving insight needed, just routine application of well-practiced techniques. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(z_2 = 2+4i\) | B1 | \(2+4i\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Product of roots \(= (2-4i)(2+4i) = 4+16 = 20\); Sum of roots \(= (2-4i)+(2+4i) = 4\) | M1 | No \(i^2\). Attempt Sum and Product of roots or Sum and discriminant |
| \(p=-4,\ q=20\): \(z^2-4z+20=0\) | A1 | Any one of \(p=-4, q=20\) |
| A1 | Both \(p=-4, q=20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((2-4i)^2 + p(2-4i) + q = 0 \Rightarrow -12-16i + p(2-4i)+q=0\) | M1 | An attempt to substitute either \(z_1\) or \(z_2\) into \(z^2+pz+q=0\), no \(i^2\) |
| Imaginary part: \(-16-4p=0\); Real part: \(-12+2p+q=0\) | ||
| \(4p=-16 \Rightarrow p=-4\); \(q=12-2p=12-2(-4)=20\) | A1 | Any one of \(p=-4, q=20\) |
| A1 | Both \(p=-4, q=20\) |
## Question 4 (Aliter):
$z^2 + pz + q = 0$, $z_1 = 2-4i$
### Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $z_2 = 2+4i$ | B1 | $2+4i$ |
### Part (a)(ii) Way 2:
| Working | Mark | Guidance |
|---------|------|----------|
| Product of roots $= (2-4i)(2+4i) = 4+16 = 20$; Sum of roots $= (2-4i)+(2+4i) = 4$ | M1 | No $i^2$. Attempt Sum and Product of roots or Sum and discriminant |
| $p=-4,\ q=20$: $z^2-4z+20=0$ | A1 | Any one of $p=-4, q=20$ |
| | A1 | Both $p=-4, q=20$ |
**Subtotal: 4 marks**
### Part (a)(ii) Way 3:
| Working | Mark | Guidance |
|---------|------|----------|
| $(2-4i)^2 + p(2-4i) + q = 0 \Rightarrow -12-16i + p(2-4i)+q=0$ | M1 | An attempt to substitute either $z_1$ or $z_2$ into $z^2+pz+q=0$, no $i^2$ |
| Imaginary part: $-16-4p=0$; Real part: $-12+2p+q=0$ | | |
| $4p=-16 \Rightarrow p=-4$; $q=12-2p=12-2(-4)=20$ | A1 | Any one of $p=-4, q=20$ |
| | A1 | Both $p=-4, q=20$ |
**Subtotal: 4 marks**
---4. Given that $2 - 4 \mathrm { i }$ is a root of the equation
$$z ^ { 2 } + p z + q = 0 ,$$
where $p$ and $q$ are real constants,
\begin{enumerate}[label=(\alph*)]
\item write down the other root of the equation,
\item find the value of $p$ and the value of $q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q4 [4]}}