Question 1188473
Sunday, April 2, 2017
7:06 PM

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A cubic function in *[tex \Large x] has exactly three complex factors of the form *[tex \Large x\ -\ z_1], *[tex \Large x\ -\ z_2], *[tex \Large x\ -\ z_3], where *[tex \Large z_i] is a zero of the function.  The family of functions given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ a(x\ -\ z_1)(x\ -\ z_2)(x\ -\ z_3)]


The particular function you are looking for is such that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(-7)\ = -8]


So solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-7\ +\ \6)(-7)(-7\ -\ 4)\ =\ -8]


For *[tex \Large a] and then find the product of your four known factors and simplify

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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