Question 1189083
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You can assume without loss of generality that *[tex \Large f(x)] has a non-zero constant term.  In the case of a polynomial function that has a zero constant term, you can repeatedly factor out *[tex \Large x] until you are left with a polynomial that <i>does</i> have a non-zero constant term or all of the possible values of *[tex \Large r] are zero which is an integer.


The Rational Zeros Theorem says that if a polynomial function with integer coefficients has a lead coefficient of *[tex \Large a_0] and a constant term coefficient of *[tex \Large a_n], and there exists a rational zero of that function, then the rational zero must have the form *[tex \Large \pm\frac{p}{q}] where *[tex \Large p] is an integer factor of *[tex \Large a_n] and *[tex \Large q] is an integer factor of *[tex \Large a_0].


Since the leading coefficient of *[tex \Large f(x)] is given to be *[tex \Large 1], if there is a rational zero for *[tex \Large f(x)] it must be an integer because in any possible zero *[tex \Large \pm\frac{p}{q}], *[tex \Large q] can only be equal to *[tex \Large 1] and since *[tex \Large p] must be an integer, *[tex \Large \pm\frac{p}{q}] must be an integer.


If *[tex \Large f(x)] has no rational zeros or if *[tex \Large a_n] is not an integer factor of *[tex \Large r], then *[tex \Large r] must be irrational because *[tex \Large r\ \in\ \mathbb{R}] and *[tex \Large \mathbb{Q}\ \small\cup\Large\ \mathbb{Q}'\ =\ \mathbb{R}]


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