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
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
But what do we do with the 4th degree factor? We use a substitution trick. Let
Et voilà! We have a quadratic that factors tidily:
So the zeros are
and
But wait! We aren't done. The problem doesn't want values of
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
