Question 197352
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And with good reason.  Using the Rational Zero Theorem, the potential rational zeros of your equation are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32]


I used an Excel spreadsheet to perform synthetic division on all 12 potential zeros and none of them worked.  That means you either have 4 real/irrational zeros, 2 real/irrational and 2 complex, or 4 complex zeros.  Using a graphing program I determined that it is the 2 and 2 situation.


The only way to solve this puppy is to use the general solution of the quartic equation.  Of course, that is a computational horror that I wouldn't wish on anybody -- even if I didn't like them very well.  Furthermore, I'm not going to attempt it here.


If you are really interested, go to http://en.wikipedia.org/wiki/Quartic_equation


Scroll down to 'Summary of Ferrari's Method' and then follow the steps outlined.  And good luck to you.  By the way, *[tex \LARGE \beta \neq 0] so this one is as difficult as it gets.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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