Question 517571
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Very, very carefully.  This trinomial is prime over the reals as evaluation of the quadratic discriminant will reveal *[tex \Large \left((-6)^2\ -\ (4)(4)(9)\ < 0\right)].  However applying the quadratic formula to the corresponding quadratic equation in Standard Form reveals that the roots are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{1,2}\ =\ \frac{3\ \pm\ 3i\sqrt{3}}{2}]


From which you can derive the following factorization of the original trinomial over the complex numbers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2x\ -\ 3\ -\ 3i\sqrt{3}\right)\left(2x\ -\ 3\ +\ 3i\sqrt{3}\right)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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