Question 637985
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If *[tex \LARGE 2\ -\ i] is a zero, then *[tex \LARGE 2\ +\ i] is also a zero because complex zeros of polynomial functions always come in conjugate pairs.


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


So now you have two factors of the original polynomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (2\ -\ i)\right)\left(x\ -\ (2\ +\ i)\right)]


Multiply them using FOIL (treating the complex quantities as single values -- remember that the product of a pair of conjugates is the difference of two squares and that *[tex \LARGE i^2\ =\ -1]


The result will be a quadratic trinomial.  Use this as the divisor of the original 4th degree polynomial using polynomial long division.  Click the link: <a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a> if you need a refresher on this process.


The quotient of the long division process will be a quadratic polynomial that is the product of the two remaining factors of the original polynomial.  Use thee quadratic formula to determine the two remaining zeros.


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