<PRE><B>Question 729978</B>
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ a^5\ +\ 6a^4\ +\ 5a^3\ =\ 0]


Factor out *[tex \LARGE a^3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ a^3\left(a^2\ +\ 6a\ +\ 5\ =\ 0\right)]


Then factor the quadratic.  You will then have 5 zeros, as you would expect for a fifth degree polynomial equation.  3 of them will be zero and the other two will be the zeros of the quadratic factor.  Hint:  5 + 1 = 6 and 5 times 1 = 5. 


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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