| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Given one complex root of cubic or quartic, find all roots |
| Difficulty | Moderate -0.3 Part (a) is straightforward substitution to verify a given root. Part (b) requires knowing that complex roots come in conjugate pairs for real polynomials, then finding the real root via polynomial division or factorization. This is a standard Further Maths technique with no novel insight required, making it slightly easier than average overall despite being Further Maths content. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \((1-2i)^2 = 1 - 2i - 2i - 4 = -3 - 4i\); \((1-2i)^3 = 1 - 6i - 12 + 8i = -11 + 2i\); \(\therefore (-11+2i) + 5(-3-4i) - 9(1-2i) + 35 = 0\) therefore \(1-2i\) is a root. | B1 M1 A1 | B0 no working. Substitution into cubic |
| Answer | Marks |
|---|---|
| Answer: If \(1 - 2i\) is a root then \(1 + 2i\) is also a root. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \((x-1+2i)(x-1-2i) = x^2 - 2x + 5\); \(x^3 + 5x^2 - 9x + 35 = (x^2-2x+5)(x+7) = 0\); Final root is \(x = -7\) | M1 A1 A1 | Unsupported answer M0 |
| Answer | Marks |
|---|---|
| Answer: Product of roots \(= -35\) OR sum \(= -5\); \((1-2i)(1+2i)a = 5a = -35\) OR \(1-2i + 1 + 2i + a = -5\); Final root is \(x = -7\) | (M1) A1 A1 |
**Part a)**
Answer: $(1-2i)^2 = 1 - 2i - 2i - 4 = -3 - 4i$; $(1-2i)^3 = 1 - 6i - 12 + 8i = -11 + 2i$; $\therefore (-11+2i) + 5(-3-4i) - 9(1-2i) + 35 = 0$ therefore $1-2i$ is a root. | B1 M1 A1 | B0 no working. Substitution into cubic
**Part b)**
Answer: If $1 - 2i$ is a root then $1 + 2i$ is also a root. | B1 |
**Method 1**
Answer: $(x-1+2i)(x-1-2i) = x^2 - 2x + 5$; $x^3 + 5x^2 - 9x + 35 = (x^2-2x+5)(x+7) = 0$; Final root is $x = -7$ | M1 A1 A1 | Unsupported answer M0
**Method 2**
Answer: Product of roots $= -35$ OR sum $= -5$; $(1-2i)(1+2i)a = 5a = -35$ OR $1-2i + 1 + 2i + a = -5$; Final root is $x = -7$ | (M1) A1 A1 |\begin{enumerate}[label=(\alph*)]
\item Show that $1 - 2\mathrm{i}$ is a root of the cubic equation $x^3 + 5x^2 - 9x + 35 = 0$. [3]
\item Find the other two roots of the equation. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q6 [7]}}