Question 676170
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If a polynomial with real coefficients has a complex zero, then the conjugate of that complex zero is also a zero.  Hence, if 2 + 3i is a zero, 2 - 3i is also a zero.


Then your factors are: *[tex \LARGE (x\ -\ 2)\left(x\ -\ (2\ +\ 3i)\right)\left(x\ -\ (2\ -\ 3i)\right)]


Multiply the three factors and then collect terms to derive your 3rd degree polynomial.  Hint for ease of computation:  Consider the complex numbers to be a single number and remember that the product of two conjugates is the difference of two squares but that since *[tex \LARGE i^2\ =\ -1] the product of two complex conjugates becomes the sum of two squares.  Write back and I'll let you know if your answer is correct. 


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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