Question 1160951
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^3\ -\ x^2\ +\ 6\ =\ (x\,-\,1)^2(2x\,+\,a)\ +\ bx\ +\ c]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ (x^2\ -\ 2x\ +\ 1)(2x\ +\ a)\ +\ bx\ + c]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ 2x^3\ -\ 4x^2\ +\ 2x\ +\ ax^2\ -\ 2ax\ +\ a\ +\ bx\ +\ c]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ 2x^3\ -\ (4\ -\ a)x^2\ +\ (2\ -\ 2a\ +\ b)x\ +\ a\ +\ c]


Equate the coefficients:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ a\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ 2a\ +\ b\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ c\ =\ 6]


Solve the 3X3 system for *[tex \Large a], *[tex \Large b], and *[tex \Large c]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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