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