Question 583689
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No.  Not answer C.


The numeric part of the answer is supposed to be the remainder that results from the synthetic division which is the same as the value of the function evaluated at the divisor.  That is, if your function is *[tex \Large f(x)] and the synthetic division divisor is *[tex \Large a] because you are testing *[tex \Large x\ -\ a] as a factor of *[tex \Large f(x)], then the rightmost number in the third row of numbers in your synthetic division is equal to *[tex \Large f(a)].  And since *[tex \Large x\ -\ a] is a factor of *[tex \Large f(x)] if and only if *[tex \Large f(a)\ =\ 0], your rightmost result is, in fact, 0, that means B is your answer.  Answers C and D are impossible because, in the case of C, the number cannot be non-zero and the answer yes, and in the case of D, the number cannot be zero and the answer no.


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