Factoring $x^5 - 5x^4 + 1$ The original problem was stated like this:
Prove that the polynomial $x^5 - 5x^4 + 1$ does not have roots of multiplicity 4.
So, factoring the polynomial would answer the question, but I don't know how to do it.
Using the root test, ${1, -1}$ should be roots, but If I divide by $x-1$ or $x+1$  $\frac {x^5 - 5x^4 + 1}{x-1}$, or $\frac {x^5 - 5x^4 + 1}{x-1}$ I get a remainder, so those are not roots?!

As suggested in the comments, try this $$\begin{align}(x-a)(x-b)^4&=(x-a)(x^4-4x^3b+6x^2b^2-4xb^3+b^4)\\&=(x^5-(a+4b)x^4+(4ab+6b^2)x^3-(4b^3+6ab^2)x^2+(b^4+4ab^3)x-ab^4)\end{align}$$
For this to match the expression $x^5-5x^4+1$, we need:
$$\begin{align}&a+4b=5\\&b(4a+6b)=0\\&b^2(4b+6a)=0\\&b^3(b+4a)=0\\&ab^4=-1
\end{align}$$
From the fifth equation, $b\neq 0$. Then from the fourth equation, $b=-4a$. Substitution into the first yields $a-16a=5\implies a=-\frac13\implies b=\frac43$. But substitution into the second equation gives $4a+6b=-\frac43+\frac{24}3\neq0$. Similarly, the third equation fails.
So the expression $x^5-5x^4+1$ cannot be factored into this form.
