Question 332608
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The rational roots theorem says that if a polynomial equation has rational roots, then these roots must be of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm\frac{p}{q}]


where *[tex \LARGE q] is an integer factor of the lead coefficient, and *[tex \LARGE p] is an integer factor of the constant term.


The factors of your lead coefficient are 1 and 2, and the factors of your constant term are 1, 2, and 4.  So take the *[tex \LARGE \pm], the 1 and 2 for the denominator, and the 1, 2, and 4 for the numerator, and make all possible distinctly different rational numbers.  Those are your <u>possible</u> rational roots.  You have a 4th degree polynomial equation, so you must have 4 roots.  Zero, 2, or 4 of them will be rational.



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