Question 467231
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*[tex \Large 2i] is two times the imaginary number *[tex \Large i], defined as *[tex \Large i^2\ =\ -1].


If the polynomial equation has the root *[tex \Large 2i], then it actually has the complex root *[tex \Large 0\ +\ 2i].  Remember that complex roots ALWAYS come in conjugate pairs, hence, if *[tex \Large 2i] is a root, then so is *[tex \Large -2i]


So the three factors of your polynomial are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x\ -\ 2i)(x\ +\ 2i)].


However, recall that the product of a pair of conjugates is the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 2i)(x\ +\ 2i)\ =\ x^2\ -\ (-4)\ =\ x^2\ +\ 4]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x^2\ +\ 4)\ =\ 0]


You can FOIL it for yourself.


Note:  This gives you the polynomial equation <i>of least degree</i> with integral coefficients that satisfies the condition that one of the roots is 3 and another root is *[tex \Large 2i].  There are an infinity of others of greater degree.


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