## Question 3:

**Either method:**

$z = 2 - 3j$ is also a root | B1 |

$(z-(2+3j))(z-(2-3j)) = ((z-2)+3j)((z-2)-3j)$ | M1 | Condone $(z+2+3j)(z+2-3j)$

$= z^2 - 4z + 13$ | A1 | Correct quadratic

$z^4 - 5z^3 + 15z^2 - 5z - 26 = (z^2 - 4z + 13)(z^2 - z - 2)$ | M1 A1 | Valid method to find other quadratic factor. Correct quadratic

$(z^2 - z - 2) = (z-2)(z+1)$

So other roots are $2$ and $-1$ | A1, A1 **[7]** | 1 mark for each root, cao

**Or method:**

$2+3j+2-3j+\gamma+\delta = 5$ | B1 | Sum of roots with substitution of roots $2\pm 3j$ for $\alpha$ and $\beta$

$(2+3j)(2-3j)\gamma\delta = -26$, $\gamma\delta = -2$ | M1 | Attempt to obtain equation in $\gamma\delta$ using a root relation and $2\pm 3j$

$\Rightarrow 4+\gamma+\delta = 5 \Rightarrow \gamma = 1-\delta$

and $13\gamma\delta = -26 \Rightarrow \gamma\delta = -2$ | M1 | Eliminating $\gamma$ or $\delta$ leading to a quadratic equation

$\Rightarrow \delta(1-\delta) = -2 \Rightarrow \delta^2 - \delta - 2 = 0$ | A1 | Correct equation obtained

$\Rightarrow (\delta+1)(\delta-2) = 0$

So other roots are $-1$ and $2$ | A1, A1 **[7]** | 1 mark for each, cao. If 2, −1 guessed from $\gamma+\delta=1$ and $\gamma\delta=-2$ give A1A1 for these equations and A1A1 for roots. **SC** factor theorem: M1 for substitution of $z=-1$ (or 2) or division by $(z+1)$ (or by $z-2$), A1 if zero obtained, B1 for root stated. M1A1A1 Max [**7/7**]. Answers only get M0M0, max [**1/7**]

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