Write the polynomial equation that has roots -2i and 1/3

Answer 1

If you want a polynomial with real coefficients, then, since -2i is a root, so is 2i..

So a possible polynomial equation would be (x - 1/3)(x - 2i)(x - (-2i)) = 0

(x - 1/3)(x² + 4 ) = 0

x³ - (1/3)x² + 4x - 4/3 = 0

Multiply both sides by 3 to get 3x³ - x² + 12x - 4 = 0

Answer 2

(x+2i)*(x-1/3) = 0. In general, a polynomial equation with roots r_1, r_2, ..., r_n will be (x-r_1) * (x-r_2) * ... * (x-r_n) = 0.