Question 1014686
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According to the Rational Roots Theorem, if there are any rational zeros, they are of the form *[tex \Large \pm \frac{p}{q}] where *[tex \Large p] is an integer factor of the constant term and *[tex \Large q] is an integer factor of the lead coefficient.


Hence the possible rational zeros are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{\pm 1,\,\pm 2,\,\pm 4,\, \pm 8,\, \pm 16\right\}]


Use Synthetic Division to test each possible zero until you have found 3 rational zeros, 1 rational zero, or no rational zeros and have exhausted all of the possibilities.  Hint: there are 3 rational zeros for this cubic equation, so if you don't find all three, check your synthetic division arithmetic.


See <a href="http://www.purplemath.com/modules/synthdiv.htm">Purple Math Synthetic  Division</a> for a refresher on Synthetic Division, if necessary.  Note that there are four pages on this topic.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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