Question 1119625
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A polynomial function of degree *[tex \Large n] has exactly *[tex \Large n] zeros counting multiplicities.  Further, complex zeros <i>always</i> appear in conjugate pairs.  If *[tex \Large a\ +\ bi] is a zero of a polynomial function, then *[tex \Large a\ -\ bi] is also a zero of that polynomial function.  And finally, if *[tex \Large \alpha] is a zero of the function, then *[tex \Large x\ -\ \alpha] is a factor of the function.


Therefore, the three factors of your 3rd degree polynomial function must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\ -\ 4)\(x\ -\ (2\ +\ i)\)\(x\ -\ (2\ -\ i)\)]


Multiply the three factors together and collect like terms to find your polynomial.  Computational hint:  Treat the two complex number factors as binomials where the complex numbers are constants (which they actually are) and recall that the product of two binomial conjugates is the difference of two squares.  However, since you are squaring *[tex \Large i] which results in -1, you actually get the SUM of two squares.  The result will be a quadratic polynomial which you can then multiply by the real number factor *[tex \Large (x\ -\ 4)].
								
								
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}}

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