Question 536558
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There exists an infinite number of polynomial functions of degree 3 that are evenly divisible by *[tex \Large x\ -\ 2].  Just select your own values for *[tex \Large \alpha] and *[tex \Large \beta], and then multiply the three binomials *[tex \Large (x\ -\ \alpha)(x\ -\ \beta)(x\ -\ 2)].  Once you have your polynomial that IS divisible by *[tex \Large x\ -\ 2], add *[tex \Large 3] to it to get a polynomial that has a remainder of 3 when divided by *[tex \Large x\ -\ 2].  Hint: *[tex \Large P(x)\ =\ (x\ -\ 2)^3\ +\ 3] will work out nicely. 


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