Question 1014617
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If *[tex \Large P\left(\frac{1}{2}\right)\ =\ 0] then *[tex \Large (2x\ -\ 1)] is a factor of *[tex \Large 2x^3\ +\ 7x^2\ +\ 6x\ -\ 5]


Use polynomial long division to divide *[tex \Large 2x^3\ +\ 7x^2\ +\ 6x\ -\ 5] by *[tex \Large (2x\ -\ 1)]


The result will be a quadratic for which you can find the zeros using the quadratic formula.


See <a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a> for a refresher on Polynomial Long Division, if necessary.


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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