Question 199415
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You are just doing the problem backwards.  In the first place, since there are two solutions, you know that you must have a second degree equation.  If you were solving a quadratic, something that looks like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2 + bx + c = 0]


You would factor the trinomial and have something that looks like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - \alpha)(x - \beta) = 0]


Then you would use the Zero Product Rule to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - \alpha = 0]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \alpha]


or

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - \beta = 0]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \beta]


Therefore, start at the end:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -3] so



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + 3 = 0]


Likewise:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 4 = 0]


Now we have our two factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x + 3)(x - 4) = 0]


So all you have to do is multiply the binomials using FOIL and you are done.


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