Not necessarily the case that all the coefficients have to be real in order for the equation to have real solutions.

Eg. (x - 2)(x - 3)(x + 5 - 4i) = 0

x³ - 4ix² + (20i - 19)x + 30 - 24i = 0

There are two real solutions but three non-real coefficients


Eg.

(x + 3i)(x + 2 - 3i)(x - 6) = 0

x³ - 4x² + (6i - 3)x - (54 + 36i) = 0

There is one real solution but two non-real coefficients.