Question 430001
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Descartes Rule of Signs:  Remembering that the lead coefficient, lacking a minus sign, is a positive coefficient, step from one term to the other counting the number of times the sign changes from + to - or - to +.


I count 1.  Therefore there is exactly 1 positive root.


If you replace x with -x, you get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(-x)\ =\ -2x^3\ +\ 5x^2\ +\ 3x\ -\ 4]


And then you count two sign changes.  Hence there are exactly 2 or 0 negative roots.


The rational roots theorem says the possible rational roots are all rational numbers of the form 


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


where *[tex \Large p] is an integer divisor of the constant term and *[tex \Large q] is an integer divisor of the lead coefficient.


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm1,\ \pm2,\ \pm4,\ \pm\frac{1}{2}]


are your possible rational roots.



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