Question 576103
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If *[tex \LARGE \alpha] is a zero of a polynomial equation, then *[tex \LARGE x\ -\ \alpha] is a factor of the polynomial.  Complex zeros always come in conjugate pairs, that is if *[tex \LARGE a\ +\ bi] is a zero, then *[tex \LARGE a\ -\ bi] is also a zero.


For your second problem, *[tex \LARGE 2i\ =\ 0\ +\ 2i] is a zero, so *[tex \LARGE 0\ -\ 2i\ =\ -2i] is also a zero.  Hence your four factors are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x\ +\ 3)(x\ +\ 2i)(x\ -\ 2i)\ =\ 0]


You can multiply out the factors for yourself.  Hint:  The product of two conjugates is the difference of two squares.  Hint #2:  Don't forget *[tex \LARGE i^2\ =\ -1] -- that will make the product of the two complex factors the SUM of two squares and eliminate the *[tex \LARGE i]s.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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