Question 624768
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Whnever you have a complex zero, you are guaranteed that the conjugate of the zero is also a zero.  The conjugate of *[tex \LARGE a\ +\ bi] is *[tex \LARGE a\ -\ bi].


So now you have two zeros for your equation, namely *[tex \LARGE 3\ -\ 4i] and *[tex \LARGE 3\ +\ 4i].  Don't forget that the product of a pair of conjugates is the difference of two squares and that *[tex \LARGE i^2\ =\ -1].  This means that the product of your conjugates is going be the difference of two squares, one of which is a negative number.


The next thing to realize is that if *[tex \LARGE \alpha] is a zero of a polynomial equation, then *[tex \LARGE x\ -\ \alpha] is a factor of the polynomial.  Furthermore the product of two or more factors of a given polynomial is itself a factor of the polynomial.


Multiply *[tex \LARGE \left(x\ -\ (3\ -\ 4i)\right)\left(x\ -\ (3\ +\ 4i)\right)]  to get a quadratic polynomial with real coefficients which is also a factor of your quartic polynomial.


Use polynomial long division to divide the original quartic by your quadratic factor.  The quotient will be another quadratic that can be solved by ordinary means (in this case it factors).


See <a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a> if you need a refresher.


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