A cubic with a complex root

Complex numbers

A-level Further Maths (9FM0) · Core Pure

ORIGINAL

Use the conjugate pair to build a real quadratic factor, then compare coefficients to find the last root.

Board and spec code confirmed against Edexcel 9FM0 · registry checked 2026-07-17How this checking works

Question

f(z) = z^3 - 5z^2 + 16z - 30

. Given that

1 + 3i

is a root of

f(z) = 0

, find the other two roots.

Every step worked, with the reasoning.

1
$1 - 3i$ is also a root.
$f(z)$ has real coefficients, so its non-real roots occur in complex-conjugate pairs.
2
$\big(z - (1 + 3i)\big)\big(z - (1 - 3i)\big) = z^2 - 2z + 10$
Sum of the pair $= 2$ and product $= (1 + 3i)(1 - 3i) = 1 - 9i^2 = 10$ , so the factor is $z^2 - (\text{sum})z + \text{product}$ .
3
$z^3 - 5z^2 + 16z - 30 = (z^2 - 2z + 10)(z - 3)$
Compare coefficients: $z \times z^2$ gives the $z^3$ term and $10 \times (-3) = -30$ gives the constant, so the linear factor is $z - 3$ . Check: $z^2$ terms $-3z^2 - 2z^2 = -5z^2$ ; $z$ terms $10z + 6z = 16z$ .
4
$z - 3 = 0 \Rightarrow z = 3$
Solve the linear factor for the real root.

Answer: $z = 1 - 3i$ and $z = 3$