Question 999816
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Complex zeros always come in conjugate pairs, so if *[tex \Large -3\ +\ i] is a zero, *[tex \Large -3\ -\ i] must also be a zero.


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


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


The only thing you need to do now is to multiply out the three factors and collect like terms.  Hint:  The product of two conjugates is the difference of two squares:  *[tex \Large (-3\ +\ i)(-3\ -\ i)\ =\ 9\ -\ (-1)\ =\ 10]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \