Find a third-degree polynomial equation with the rational coefficients that has roots - 3 and 1+i

x³ + x² - 4x + 6 = 0

Step-by-step explanation:

Imaginary roots come in conjugate pairs. So if 1+i is a root, then 1-i is also a root.

(x - (-3)) (x - (1+i) (x - (1-i)) = 0

(x + 3) (x² - (1+i) x - (1-i) x + (1+i) (1-i)) = 0

(x + 3) (x² - x - ix - x + ix + 1 - i²) = 0

(x + 3) (x² - 2x + 2) = 0

x (x² - 2x + 2) + 3 (x² - 2x + 2) = 0

x³ - 2x² + 2x + 3x² - 6x + 6 = 0

x³ + x² - 4x + 6 = 0