Question

Find a third-degree polynomial equation with rational coefficients that has the given numbers as roots. 1 and 3i

Answer 1

Answer:

x³ − x² + 9x − 9 = 0

Step-by-step explanation:

Imaginary roots come in conjugate pairs.  So if 3i is a root, then -3i is also a root.

(x − 1) (x − 3i) (x − (-3i)) = 0

(x − 1) (x − 3i) (x + 3i) = 0

(x − 1) (x² − 9i²) = 0

(x − 1) (x² + 9) = 0

x (x² + 9) − (x² + 9) = 0

x³ + 9x − x² − 9 = 0

x³ − x² + 9x − 9 = 0