Question 662478: Iam trying to find all the zeros for
x5+6x4-x3-6x2-20x-120 found -6
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
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  , 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:
-6 | 1 6 -1 -6 -20 -120
-6 0 6 0 120
1 0 -1 0 20 0
From this we can determine that and  are factors of  , 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  and substitute into :
Et voilą! We have a quadratic that factors tidily:
(u\ +\ 4))
So the zeros are
and
But wait! We aren't done. The problem doesn't want values of  that make the original polynomial equal to zero, it wants values of  . So reverse the substitution, thus:
hence
OR
hence
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
My calculator said it, I believe it, that settles it
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