| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Find conjugate roots from polynomial |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge that complex roots come in conjugate pairs, followed by polynomial multiplication to find unknown coefficients. While the concept is standard for FM students, the algebraic manipulation of expanding two quadratics and equating coefficients requires careful multi-step work, placing it moderately above average difficulty. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies \(1 + 3i\) as a second root of the quartic equation | B1 | |
| \(z = 1 + 3i\) is another root. \(z^2 - 2z + 10\) is a factor. \((z^2 - 2z + 10)(z^2 + bz + 8) \equiv z^4 - 2z^3 + pz^2 + rz + 80\) | M1 | Uses a pair of conjugate roots to find a quadratic factor |
| Comparing \(z^3\)-terms gives \(b = 0\) | A1 | Finds one correct quadratic factor |
| \(\therefore\) the quartic is \((z^2 - 2z + 10)(z^2 + 8)\) | A1 | Correctly expresses quartic as product of two quadratic factors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z^4 - 2z^3 + 18z^2 - 16z + 80 = 0\) | M1 | Substitutes \(1 + 3i\) or \(1 - 3i\) or one of their roots from factorisation in (a) into the quartic equation and compares Re and Im parts, or compares coefficients of \(z^2\) and \(z\) from the (possibly partial) expansion of their product of quadratics with the given quartic |
| \(p = 18\), \(r = -16\) | A1 | Finds the correct values of \(p = 18\) and \(r = -16\) |
# Question 4:
## Part 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies $1 + 3i$ as a second root of the quartic equation | B1 | |
| $z = 1 + 3i$ is another root. $z^2 - 2z + 10$ is a factor. $(z^2 - 2z + 10)(z^2 + bz + 8) \equiv z^4 - 2z^3 + pz^2 + rz + 80$ | M1 | Uses a pair of conjugate roots to find a quadratic factor |
| Comparing $z^3$-terms gives $b = 0$ | A1 | Finds one correct quadratic factor |
| $\therefore$ the quartic is $(z^2 - 2z + 10)(z^2 + 8)$ | A1 | Correctly expresses quartic as product of two quadratic factors |
## Part 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z^4 - 2z^3 + 18z^2 - 16z + 80 = 0$ | M1 | Substitutes $1 + 3i$ or $1 - 3i$ or one of their roots from factorisation in (a) into the quartic equation and compares Re and Im parts, or compares coefficients of $z^2$ and $z$ from the (possibly partial) expansion of their product of quadratics with the given quartic |
| $p = 18$, $r = -16$ | A1 | Finds the correct values of $p = 18$ and $r = -16$ |
---4
\begin{enumerate}[label=(\alph*)]
\item Express $z ^ { 4 } - 2 z ^ { 3 } + p z ^ { 2 } + r z + 80$ as the product of two quadratic factors with real coefficients.\\
[4 marks]\\
4 It is given that $1 - 3 \mathrm { i }$ is one root of the quartic equation\\
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4
\item Find the value of $p$ and the value of $r$.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q4 [6]}}