Paper 1 · Complex numbers

The equation $x^3 - 2x^2 + 4x - 8 = 0$ has a root $x = 2i$ . Find the other two roots.

x = -2i

and

x = -2

x = 2i

and

x = 2

x = -2i

and

x = 4

x = -2i

and

x = 2

Explanation: Complex roots occur in conjugate pairs, so

x=-2i

is a root and

x^2+4

is a factor. Then

x^3-2x^2+4x-8=(x-2)(x^2+4)

, giving the third root

x=2

Derived from ZIMSEC Mathematics Paper 1, November 2018, Q1

Get the full paper, not just one question

Danho has every sitting for this paper, with your progress tracked question by question, offline.