Question 378418
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First use the Rational Roots Theorem which says that if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha_0x^n\ +\ \alpha_1x^{n-1}\ +\ \cdots\ +\ \alpha_{n-1}x\ +\ \alpha_n\ =\ 0]


has a rational root then it must be of the form:


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


Where *[tex \Large p] is an integer factor of *[tex \Large \alpha_n] and *[tex \Large q] is an integer factor of *[tex \Large \alpha_0]


Using that, the only possible rational factors of


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^3\ +\ 9x^2\ -\ 8x\ -\ 15]


are


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\,\pm\,1,\,x\,\pm\,3,\,x\,\pm\,5,\,x\,\pm\,15,\,x\,\pm\,\frac{1}{2},\,x\,\pm\,\frac{3}{2},\,x\,\pm\,\frac{5}{2},\,x\,\pm\,\frac{15}{2}]


Use synthetic division to test these 16 trial divisors until you either find one that divides the original cubic polynomial evenly, leaving you with the successful trial divisor as one factor and a quotient consisting of a quadratic trinomial as another (and hopefully further factorable) factor, or you exhaust all of the possibilities and thereby determine that your denominator is prime.


If you need a refresher on Synthetic Division, see:


<a href="http://www.purplemath.com/modules/synthdiv.htm">Purple Math: Synthetic Division</a> (There are 4 pages -- read them all).


Write back and let me know how you make out.


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