Question 1119660
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You have done all of the work by factoring the quintic equation.  Just set each of the factors equal to zero and solve.  You have a linear binomial so one of the roots is x = -1.  Solve the other two with the quadratic formula.  Note that four of your roots are complex.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2y^2\ +\ 2y^2z^2\ +\ 2x^2z^2\ -\ x^4\ -\ y^4\ -\ z^4\ =\ -(x\ -\ y\ -\ z)(x\ +\ y\ -\ z)(x\ -\ y\ +\ z)(x\ +\ y\ +\ z)]
								
								
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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