Question 221983
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Missed it by <i>THAT</i> much.


Factoring out the *[tex \Large t^2] was certainly the right thing to do, but if you notice, the numerical coefficients on the two original terms have the factor 2 in common.  Hence, the correct first step would have been to factor out *[tex \Large 2t^2] leaving you with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2t^2(y^4-16)]


But you aren't done, not by a long shot.


Notice that *[tex \Large y^4-16] is the difference of two squares, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2t^2(y^2-4)(y^2+4)]


And then notice that *[tex \Large y^2-4] is also the difference of two squares, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2t^2(y-2)(y+2)(y^2+4)]


And that's as far as you can go presuming you are confining yourself to the real numbers.


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