# Question 8:

## Part (i):

$3(1+3j)^3 - 2(1+3j)^2 + 22(1+3j) + 40$ | M1 | Substitute $z = 1+3j$ into cubic

$= 3(-26-18j) - 2(-8+6j) + 22(1+3j) + 40$ | A1, A1 | $(1+3j)^2 = -8+6j$, $(1+3j)^3 = -26-18j$

$= (-78+16+22+40) + (-54-12+66)j = 0$ | A1 | Simplification (correct) to show this comes to $0$ and so $z = 1+3j$ is a root

**[4]**

## Part (ii):

All cubics have 3 roots. As the coefficients are real, the complex conjugate is also a root. This leaves the third root, which must therefore be real. | E1 | Convincing explanation

**[1]**

## Part (iii):

$1-3j$ must also be a root | B1 |

Sum of roots $= -\frac{-2}{3} = \frac{2}{3}$ **OR** product of roots $= -\frac{40}{3}$ **OR** $\sum\alpha\beta = \frac{22}{3}$ | M1 | Attempt to use one of $\sum\alpha,\ \alpha\beta\gamma,\ \sum\alpha\beta$

$(1+3j)+(1-3j)+\alpha = \frac{2}{3}$ **OR** $(1+3j)(1-3j)\alpha = -\frac{40}{3}$ **OR** $(1-3j)(1+3j)+(1-3j)\alpha+(1+3j)\alpha = \frac{22}{3}$ | A2,1,0 | Correct equation

$\Rightarrow \alpha = -\frac{4}{3}$ is the real root | A1 | Cao

**OR**

$1-3j$ must also be a root | B1 |

$(z-1+3j)(z-1-3j) = z^2-2z+10$ | M1, A1 | Use of factors; Correct quadratic factor

$3z^3-2z^2+22z+40 \equiv (z^2-2z+10)(3z+4) = 0$ | A1, A1 | Correct linear factor (by inspection or division); Cao

$\Rightarrow z = -\frac{4}{3}$ is the real root

**[5]**

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