Question 708810
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Let *[tex \LARGE u\ =\ h^2].  Then *[tex \LARGE u^2\ -\ 256] is equivalent to *[tex \LARGE h^4\ -\ 256].


Factor the difference of two squares:  *[tex \LARGE u^2\ -\ 256\ =\ (u\ -\ 16)(u\ +\ 16)]


Then substitute back:  *[tex \LARGE (h^2\ -\ 16)(h^2\ +\ 16)]


But then you still have a difference of two squares factor, so factor that: *[tex \LARGE (h\ -\ 4)(h\ +\ 4)(h^2\ +\ 16)]


If you are factoring over the reals, then you are done.  However if you are factoring over the complex numbers, you can still factor the sum of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h^4\ -\ 256\ =\ (h\ -\ 4)(h\ +\ 4)(h\ -\ 4i)(h\ +\ 4i)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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