Question 662478
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Having found -6 is a good start, and if you continued looking for rational roots from there, you were doomed to failure; there aren't any.  The other four zeros consist of a pair of irrationals and a conjugate pair of complex numbers.  The thing is, after having found the factor *[tex \LARGE x\ +\ 6], you were staring at a 4th degree polynomial that we now know has no rational factors.  Woe is me!  What to do now?


Fortunately, this particular 4th degree polynomial is so configured that you can actually treat it like a quadratic.  Read on.


First let's do the synthetic division with -6:


<pre>

-6  |    1    6   -1   -6  -20 -120
    <u>         -6    0    6    0  120</u>
         1    0   -1    0   20    0

</pre>

From this we can determine that *[tex \LARGE x\ +\ 6] and *[tex \LARGE x^4\ -\ x^2\ -\ 20] are factors of *[tex \LARGE x^5\ +\ 6x^4\ -\ x^3\ -\ 6x^2\ -\ 20x\ -\ 120], and that -6 is indeed a real rational zero of the 5th degree polynomial.


But what do we do with the 4th degree factor?  We use a substitution trick.  Let *[tex \LARGE u\ =\ x^2] and substitute into *[tex \LARGE x^4\ -\ x^2\ -\ 20]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ u\ -\ 20]


Et voilą! We have a quadratic that factors tidily:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ -\ 5)(u\ +\ 4)]


So the zeros are


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 5]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -4]


But wait!  We aren't done.  The problem doesn't want values of *[tex \LARGE u] that make the original polynomial equal to zero, it wants values of *[tex \LARGE x]. So reverse the substitution, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 5]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{5}]


OR


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ -4]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm{2i}]


And we are done.  We started with a 5th degree polynomial and found 5 zeros satisfying the Fundamental Theorem of Algebra.  Time to sit around in a circle, hold hands, and sing "Kumbaya."


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