Question 1002792
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If a polynomial has a complex zero, then the conjugate of that zero is also a zero of the polynomial.  Since *[tex \Large 0\ -\ 3i] is a zero, *[tex \Large 0\ +\ 3i] must also be a zero.


If *[tex \Large \alpha] is a zero of a polynomial, then *[tex \Large x\ -\ \alpha] is a factor of the polynomial.  Hence, *[tex \Large x\ -\ 3i] and *[tex \Large x\ +\ 3i] are both factors of the given function.


Multiply these two factors (Hint:  The product of two conjugates is the difference of two squares, but since *[tex \Large i^2\ =\ -1] you will get the sum of two squares.


Use Polynomial Long Division to divide the above product into the original function.  Since the divisor is quadratic and the dividend is quartic, the quotient will be quadratic and, therefore, solvable by any convenient means. Hint: it factors, but the quadratic formula is quicker.


If you need a refresher on Polynomial Long Division, see: <a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a>.  Note that the divisor is a quadratic with no linear term, so you will need a placeholder in the divisor.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
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