Question 415382
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If a polynomial equation has a complex root, then the conjugate of that complex root is also a root.


Hence, your polynomial has three roots and must be at least degree 3.


The problem here is that I'm not completely certain that you expressed the complex root correctly.  What you said was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ +\ \sqrt{3i}]


Which, if you actually calculate *[tex \Large \sqrt{i}], turns out to be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\ +\ \sqrt{6}}{2}\ +\ \frac{\sqrt{6}}{2}i]


And I don't think that is what you meant at all.  What I think you were trying to say was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ +\ i\sqrt{3}]


Given that, your complex conjugate is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ i\sqrt{3}]


Hence the three factors are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ 3\right)\left(x\ -\ 2\ -\ i\sqrt{3}\right)\left(x\ -\ 2\ +\ i\sqrt{3}\right)]


Now all that is left is to consider each of the complex factors as a binomial and multiply the three binomials together.  After you have collected terms and set the whole thing equal to zero, you will have your desired polynomial.  Hint: Remember that a pair of conjugates multiplied results in the difference of two squares.  Another hint:  Don't forget that *[tex \Large i^2\ =\ -1]


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