Question 312825
<font face="Garamond" size="+2">


First apply the Rational Root Theorem.  The possible rational roots of:


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


are any rational number of the form


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


where *[tex \LARGE q\ \in\ \mathbb{Z}] and *[tex \LARGE q] is a factor of *[tex \LARGE \alpha_n] and  *[tex \LARGE p\ \in\ \mathbb{Z}] and *[tex \LARGE p] is a factor of *[tex \LARGE \alpha_0] 


Your lead coefficient being 1 simplifies things a little...your possible rational roots are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm1,\,\pm3,\,\pm5,\,\pm15]


Use Synthetic Division and the Remainder Theorem to determine if any of these 8 possibilities are actually roots of the given equation.


If you are unfamiliar with the process of synthetic division, check out Purple Math's explanation at http://www.purplemath.com/modules/synthdiv.htm.


The Remainder Theorem says that the remainder when you use synthetic division with a divisor of *[tex \LARGE a] is equal to *[tex \LARGE f(a)], and if *[tex \LARGE f(a)\ =\ 0], then *[tex \LARGE a] must be a root of *[tex \LARGE f]


Fortunately for this particular problem, one of the rational roots actually works.  When you find the correct synthetic divisor, you will be left with the coefficients of a quadratic equation that can be solved with the quadratic formula to yield a conjugate pair of complex roots.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>