<PRE><B>Question 690752</B>
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If 6 - 5i is a zero, then 6 + 5i must also be a zero.  Therefore both [x - (6 - 5i)] and [x - (6 + 5i)] are factors of the given polynomial function.


Multiply the two factors to get x&lt;sup&gt;2&lt;/sup&gt; - 12x + 61, which, since it is a composite of two factors of the given polynomial must also be a factor of the given polynomial.


Use polynomial long division to find the other quadratic factor of the given quartic:

                              x&lt;sup&gt;2&lt;/sup&gt; -   6x -   55
              ---------------------------------
x&lt;sup&gt;2&lt;/sup&gt; - 12x + 61 | x&lt;sup&gt;4&lt;/sup&gt; - 18x&lt;sup&gt;3&lt;/sup&gt; + 78x&lt;sup&gt;2&lt;/sup&gt; + 294x - 3355
                x&lt;sup&gt;4&lt;/sup&gt; - 12x&lt;sup&gt;3&lt;/sup&gt; + 61x&lt;sup&gt;2&lt;/sup&gt;
              -------------------------
                   -  6x&lt;sup&gt;3&lt;/sup&gt; + 17x&lt;sup&gt;2&lt;/sup&gt; + 294x
                   -  6x&lt;sup&gt;3&lt;/sup&gt; + 72x&lt;sup&gt;2&lt;/sup&gt; - 366x
                   ---------------------------
                          - 55x&lt;sup&gt;2&lt;/sup&gt; + 660x - 3355
                          - 55x&lt;sup&gt;2&lt;/sup&gt; + 660x - 3355
                         ---------------------
                                             0

Finally, factor  x&lt;sup&gt;2&lt;/sup&gt; - 6x - 55 = (x + 5)(x - 11) from which you can get your final two zeros.


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John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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