Question 1159703
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Complex zeros always appear in conjugate pairs, therefore if *[tex \Large 2i] is a zero, *[tex \Large -2i] must also be a zero.  Real zeros have no such restriction, so the simplest polynomial with the given zeros has exactly three zeros, and is a 3rd-degree polynomial.


If *[tex \Large a] is a zero of a polynomial function then *[tex \Large x\ -\ a] is a factor of the polynomial.  Hence, your desired polynomial function, in factored form is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ (x\,+\,3)(x\,+\,2i)(x\,-\,2i)]


While the above is incontrovertibly a correct answer to the question, I suspect that you will need to present the answer in standard form, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(x)\ =\ Ax^3\ +\ Bx^2\ +\ Cx\ +\ D]


And this will require that you multiply the three factors together and collect like terms.  Hint: The product of two conjugate binomials is the difference of two squares.  Don't forget that *[tex \Large i^2\ =\ -1].  I'll leave the rest in your capable hands.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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