Question

Solve the cubic equation to find the three *complex number roots* (ie in terms of a+bi).

3z³-z²+12z-4=0

Answer 1

z = 1/3+0i, z = 0+2i, z = 0-2i

Explanation:

3z³-z²+12z-4=0

(3z-1)(z²+4)=0

3z-1=0 and z²+4=0

z = 1/3 and z=±2i

z = 1/3+0i, z = 0+2i, z = 0-2i