Question 260990
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Well, one of your roots is obviously *[tex \Large t\ =\ 0].  Now you can then say that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^4\ =\ 256]


Let *[tex \Large t^2\ = x]


Then you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 256]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 16]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ = -16]


but since *[tex \Large x\ =\ t^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ =\ 16]


which means that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm4]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ =\ -16]


which means that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm4i]


Giving us the expected 5 roots for the initial quintic equation.


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