Question 314984
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Let *[tex \Large u\ =\ y^2] and substitute *[tex \Large u^2] for *[tex \Large y^4]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25u^2\ -\ 36]


Now you have the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5u\ -\ 6)(5u\ +\ 6)]


Replace *[tex \Large u] with *[tex \Large y^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5y^2\ -\ 6)(5y^2\ +\ 6)]


If you really meant to say "factor completely over the the rationals" then you are done, otherwise:


The left-hand binomial can be factored as the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\sqrt{5}\ -\ \sqrt{6})(y\sqrt{5}\ +\ \sqrt{6})(5y^2\ +\ 6)]


If you really meant "factor completely over the reals" then you are done, otherwise, you can factor the other binomial involving *[tex \Large y^2] as the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\sqrt{5}\ -\ \sqrt{6})(y\sqrt{5}\ +\ \sqrt{6})(y\sqrt{5}\ -\ i\sqrt{6})(y\sqrt{5}\ +\ i\sqrt{6})]



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