## Find a polynomial function of least degree having only real coefficients, a leading of 1, and zeros of 2 and 2+i.

Question

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## Answers ( )

Answer:y = x³ − 6x² + 13x − 10

Step-by-step explanation:Complex roots come in pairs, so if 2 + i is a root, then 2 − i is also a root.

y = (x − 2) (x − (2 + i)) (x − (2 − i))

y = (x − 2) (x − 2 − i) (x − 2 + i)

y = (x − 2) ((x − 2)² − i²)

y = (x − 2) (x² − 4x + 5)

y = x (x² − 4x + 5) − 2 (x² − 4x + 5)

y = x³ − 4x² + 5x − 2x² + 8x − 10

y = x³ − 6x² + 13x − 10