Question 1163497
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Complex roots always appear in conjugate pairs.  So if *[tex \Large 2\,+\,i] is a root of the equation, so is *[tex \Large 2\,+\,i]


Therefore both *[tex \Large \(x\ -\ (2\,+\,i)\)] and *[tex \Large \(x\ -\ (2\,-\,i)\)] are factors of the cubic polynomial.  I leave it as an exercise for the student to verify that the product of these two factors is *[tex \Large x^2\ -\ 4x\ +\ 5]


Using Polynomial Long Division


*[illustration polylongdiv.jpg]


In order for *[tex \Large x^2\ -\ 4x\ +\ 5] to be a factor of the original cubic polynomial, the following facts must hold:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (b\,-\,5)\ +\ 4(a\,+\,4)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ 5(a\ +\ 4)\ =\ 0]


Solve the 2X2 system for *[tex \Large a] and *[tex \Large b]
								
								
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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