Question 549764
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You are only supposed to put one problem per post.  So what I will do is tell you how to solve all problems of this type instead of solving these specific ones.


First:  A polynomial function of degree *[tex \Large n] ALWAYS has *[tex \Large n] roots.


Second:  Irrational and complex roots ALWAYS come in conjugate pairs.  That is, if *[tex \Large a\ +\ b\sqrt{c}] where *[tex \Large a], *[tex \Large b], and *[tex \Large c] are real numbers, *[tex \Large c\ >\ 0], and *[tex \Large c] is not a perfect square, then *[tex \Large a\ -\ b\sqrt{c}] is also a root.  Also, if *[tex \Large a\ +\ bi] is a root, then *[tex \Large a\ -\ bi] is also a root.


Third:  If *[tex \Large \alpha] is a root, then *[tex \Large x\ -\ \alpha] is a factor of the polynomial.  So if the roots of an *[tex \Large n] degree polynomial are the set *[tex \Large \left\{\alpha_1,\,\alpha_2,\,\cdots\,\,\alpha_n\right\}], then the completely factored polynomial is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \left(x\ -\ \alpha_1\right)\left(x\ -\ \alpha_2\right)\,\cdots\,\left(x\ -\ \alpha_n\right)]


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