Question 203020
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If you think you are going to get someone to do your entire homework assignment for you, let me disabuse you of such a notion right now.  I'll do one, and you can do the rest.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x+3i)(x-3i)(x-2)(x+1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x+3i)(x-3i)] is a conjugate pair, so the product is the difference of two squares (remembering that *[tex \Large i^2 = -1]):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9]


Multiply the other two factors using FOIL:


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


Now multiply the two partial products:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x^2 + 9)(x^2 - x - 2) = x^4 - x^3 - 2x^2 + 9x^2 - 9x - 18 = x^4 - x^3 + 7x^2 - 9x - 18]


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