Question 1038818
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Complex roots always appear in conjugate pairs, so the simplest polynomial function with the given zeros also has *[tex \Large 1\ -\ 2i] as a zero.  Hence, the simplest polynomial with the given roots is one of the third degree.


Therefore, your polynomial function has the following factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ (x\ +\ 4)\left(x\ -\ (1\ +\ 2i)\right)\left(x\ -\ (1\ -\ 2i)\right)]


You can expand it for yourself.  Hint:  The product of two complex conjugates is the SUM of two squares.


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 \ \ \ \ \ \ \ \ \ \  

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