Find the cubic equation that has −1 and 2i as roots.

Explanation:

Assuming the coefficients are real numbers, this means 2i pairs up with -2i. They are conjugate pair roots. So we have these three roots

-1, 2i, -2i

Meaning that

x = -1, x = 2i, x = -2i

x+1 = 0,  x^2 = -4 see note below

x+1 = 0 or x^2+4 = 0

(x+1)(x^2+4) = 0

x(x^2+4)+1(x^2+4) = 0

x^3+4x+x^2+4 = 0

x^3+x^2+4x+4 = 0

To check the answer, plug each root (-1, 2i and -2i) into this equation. You should get 0 on the left side each time.

note: If we have x = 2i, then we can square both sides to get x^2 = 4i^2 which becomes x^2 = -4. The same happens to x = -2i. You should find that solving x^2 = -4 leads back to the two roots x = 2i or x = -2i.