Question 332383
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^3\ -\ 7x^2\ -\ 128x\ +\ 448\ =\ 0]


First, the hard way.


Step 1:  Use the Rational Roots Theorem to determine all of the <i><b>possible </b></i> rational roots to your equation.  If you don't know the Rational Roots Theorem, there is a very good discussion here:


<a href="http://en.wikipedia.org/wiki/Rational_root_theorem">Wikipedia Rational Roots Theorem</a>


Once you have the list of potential rational number roots, you need to start testing them.  Fortunately, since you are working with a cubic, you only need to find one and the results of your successful test will leave you with a quadratic that you can solve (in this case by factoring).  Unfortunately, the Rational Roots Theorem is going to identify a huge boatload of potential roots, most of which won't work.


If you have or can get your hands on a graphing calculator, you can save yourself a TON of work at this point.  Graph the function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 2x^3\ -\ 7x^2\ -\ 128x\ +\ 448]


and see if you can find a place where it looks like it might cross the x-axis at a whole number.  Try that whole number first.


The best test to use when you are trying a potential zero is Synthetic Division.  If you aren't familiar with Synthetic Division, try this:


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


Have patience and go through all 4 pages of it.  It will be worth your while to know this someday.


Once you get a Zero remainder result meaning you have found one of the zeros of this cubic equation, the quotient part of the synthetic division will be the coefficients of an easily factorable quadratic equation.  Solve the quadratic for your other two roots.


On the other hand, if all of that seems like too much work.  Try:


<a href="http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=equations&s2=solve&s3=basic">Quick Math Equation Solver</a>


Just type in your equation, exactly the way you put it in on this site, into the "Solve" box in the center of the page, then click the "Solve" button.  QuickMath will do the rest.


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