Question 1126935
<font face="Times New Roman" size="+2">


Complex zeros always appear in conjugate pairs, thus if *[tex \Large a\ +\ bi] is a zero, then *[tex \Large a\ -\ bi] is also a zero.  Hence the 5 zeros of your desired polynomial are: *[tex \Large 0\ +\ 2i,\ 0\ -\ 2i,\ 1\ -\ 2i,\ 1\ +\ 2i,\ \ ] and *[tex \Large -5].


If *[tex \Large \alpha] is a zero of a polynomial function *[tex \Large \rho(x)], then *[tex \Large x\ -\ \alpha] is a factor of the polynomial.  So, in factored form, your polynomial is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ (x\ +\ 2i)(x\ -\ 2i)(x\ -\ 1\ -\ 2i)(x\ -\ 1\ +\ 2i)(x\ +\ 5)]


Multiplying it out is left as an exercise for the student.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>