Question 611040
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If -6 is a zero, then *[tex \LARGE x\ +\ 6] is a factor of the polynomial.


if *[tex \Large i] is a zero, then *[tex \Large -i] is also a zero because complex roots always appear in conjugate pairs, i.e. *[tex \Large a\ \pm\ bi]


The product of a pair of conjugates is the difference of two squares, so the product of *[tex \Large (x\ +\ i)(x\ -\ i)] is *[tex \LARGE x^2\ +\ 1]


So the product


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ k(x\ +\ 6)(x^2\ +\ 1)]


is the family of polynomial functions contained in which is the particular function you are looking for.


Expand the product above and set the whole thing equal to 60.  Then substitute -3 for x, leaving you with a single variable equation in *[tex \Large k].  Solve for *[tex \Large k] and collect terms to write your specific third degree polynomial.


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