Question 274328
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It doesn't factor over the rationals.  Check the discriminant, that is *[tex \Large b^2\ -\ 4ac].  Works out to 612.  Not a perfect square.  In fact, the best you can do is *[tex \Large 2\sqrt{153}]


Exact solution set:  *[tex \Large x\ =\ -5\ \pm\ \sqrt{153}] using the quadratic formula.


Of course the trinomial does factor -- it is just uglier than a mud fence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 10x\ -\ 128\ =\ \left(x\ +\ 5\ +\ \sqrt{153}\right)\left(x\ +\ 5\ -\ \sqrt{153}\right)]



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