Question 497061
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Re-read the instructions for posting on this site, especially the instruction that says "One question per post."  What is NOT going to happen is for someone to do your whole homework assignment, or worse, your test, for you.


Here are a couple of rules to help you though.  Rational zeros of a polynomial function, if they exist, must be of the form *[tex \Large \pm\frac{p}{q}] where *[tex \Large \pm{p}] is an integer factor of the constant term and *[tex \Large \pm{q}] is an integer factor of the lead coefficient.


A polynomial function of degree *[tex \Large n] has exactly *[tex \Large n] zeros, some of which may be complex.   Complex zeros ALWAYS show up as conjugate pairs, that is, if *[tex \Large a\ +\ bi] is a zero, then *[tex \Large a\ -\ bi] must also be a zero.


A corollary to the above is that polynomial functions of odd degree must have at least one real zero.


Using the first rule I gave you, you can use either polynomial long division or synthetic division to determine if a potential rational zero is actually a zero of the given polynomial.  The quotient of a successful division that identifies a zero can be couched as a lesser degree polynomial funcition.  Continue the process until you cannot find any more rational roots or you end up with a quadratic quotient that you can solve with the quadratic formula.  In fact, if you get down to a quartic or a cubic, there are general formulas for these too, but you don't have any where you will need to go to those lengths.


If you are a little rusty on your poly long division or synthetic division, check out:


<a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a>


<a href="http://www.purplemath.com/modules/synthdiv.htm">Purple Math Synthetic Division</a>


To determine the number of real zeros use Descartes' Rule of Signs


<a href="http://www.purplemath.com/modules/drofsign.htm">Purple Math Descartes' Rule of Signs</a>


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