Question 474207
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You can, if such nonsense appeals to you, muck about with trial and error trying to find sets of factors of 6 and 10 so that the middle term comes out to be 19.  But there is a much easier way.


Set the trinomial equal to zero and then solve using the quadratic formula (alternatively, use one of the many on-line quadratic solvers).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-19\ \pm\ \sqrt{361\ -\ 240}}{12}]


If you get to this point and you have a radicand that is not a perfect square, you can just say that the trinomial is prime and quit here.  Your example is the opposite situation, so continue:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{30}{12}\ =\ -\frac{5}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{8}{12}\ =\ -\frac{2}{3}].


So if *[tex \Large x\ =\ -\frac{5}{2}] then *[tex \Large 2x\ +\ 5\ =\ 0]


Likewise, if *[tex \Large x\ =\ -\frac{2}{3}] then *[tex \Large 3x\ +\ 2\ =\ 0]


And there you have your two factors.


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