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

it is true

step-by-step explanation:

all the combinations add up i used   calculator