Question 858768
<font face="Times New Roman" size="+2">


Complex zeros always come in conjugate pairs, that is, if *[tex \Large \alpha\ +\ \beta{i}] is a zero of a polynomial then *[tex \Large \alpha\ -\ \beta{i}] is also a zero of that polynomial.


Since your given zeros are 2 and *[tex \Large -3i] (which is to say *[tex \Large 0\ -\ 3i]), the third and final zero must be *[tex \Large 0\ +\ 3i], or simply *[tex \Large 3i]


If *[tex \Large \phi] is a zero of a polynomial, then *[tex \Large x\ -\ \phi] is a linear factor of the polynomial.  Hence the linear factors of the desired polynomial are *[tex \Large \left(x\ -\ 2\right)\left(x\ -\ (-3i)\right)\left(x\ -\ 3i\right)], or more simply:  *[tex \Large \left(x\ -\ 2\right)\left(x\ +\ 3i\right)\left(x\ -\ 3i\right)]


You can multiply the factors to derive the standard form polynomial yourself.  Hint:  do the two complex factors first, remembering that the product of a pair of conjugates is the difference of two squares and that *[tex \Large i^2\ =\ -1]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>