Question 199795
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If *[tex \Large x = 0], or *[tex \Large x = 1], or *[tex \Large x = 2], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 0], or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 1 = 0], or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 2 = 0]


Since each of these three expressions can be equal to zero depending on the value of *[tex \Large x], their product must also be zero if *[tex \Large x] is equal to any of the three possible solutions, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x - 1)(x - 2) = 0]


Now, that is "an" equation that satisfies the requirements of the problem.  However, I suspect that the desired answer is a 3rd degree polynomial in *[tex \Large x] which can be obtained by first multiplying the two binomials using FOIL and then multiplying the result by *[tex \Large x]


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