Question 699319
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Both complex zeros and irrational zeros always appear in conjugate pairs.  Hence, if *[tex \LARGE 2\ -\ i] is a zero, then *[tex \LARGE 2\ +\ i] is also a zero.  If *[tex \LARGE 1\ +\ \sqrt{3}] is a zero, then *[tex \LARGE 1\ -\ sqrt{3}] is also a zero.


If *[tex \LARGE \alpha] is a zero of a polynomial equation, then *[tex \LARGE x\ -\ \alpha] must be a factor of the polynomial.


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ (2\ -\ i))(x\ -\ (2\ +\ i))(x\ -\ (1\ +\ \sqrt{3}))(x\ -\ (1\ -\ \sqrt{3}))]


is the fully factored polynomial.  Multiply out the factors and set the resulting 4th degree polynomial equal to zero.  Hint:  Multiply the first two binomial factors then the last two binomial factors.  Treat the inner parentheticals as a single term.  Remember that the product of a pair of real number conjugates is the difference of two squares, and the product of a pair of complex conjugates is the sum of two squares.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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