Can completing the square apply to more higher degree? I have  some trouble about completing the square lessons.
As we know , minimum value of  $\ \ 3x^{2} + 4x + 1   $ is $-\frac{1}{3}$ when $x=-\frac{2}{3}$ by completing the square.
Then , how about minimum value of $\ \ x^5 +4x^4 + 3x^3 +2x^2 + x + 1$ when  $(-2<x<2)$ ?
According to Wolfram Alpha , minimum value is $12 - 5\sqrt{5}$ when $x=-\frac{3}{2} + \frac{\sqrt{5}}{2}$
I try to put 5th degree to form $(x-a)^5$ or any nth degree but it doesn't work.
I'm just curious how  completing the square can apply to higher deree equations to find their minimum or maximum value.  Thank you for every help comments.

We would generally use the derivative for this. Set it to zero and find the extrema. Via this you can probably find some abstract factorisation that includes these extrema, but it wouod be hard to interpret I think. This is an interesting idea. I think I learnt that you cannot factorise up to order 5, but anything under that you can :)  But these extrema won't all be the same value so why factorise it as $ (x-a)^{5} $
