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?!
 A: 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.
A: Factoring is not a good idea; factoring a 5th degree polynomial is difficult at best, and usually impossible.
You can show this using derivatives. If $a$ is a repeated root of $p(x)$ with multiplicity 4, then $p(x) = (x-a)^{4}(x-b)$.  If you look at the first three derivatives, you can see that $p(a)$, $p^{\prime}(a)$, $p^{\prime\prime}(a)$, and $p^{(3)}(a)$ all should evaluate to $0$.  If you look at your polynomial, you can see that this is not the case (the roots are easy to find for all of the derivatives).
A: I am surprised nobody offered this approach yet: If a polynomial $f \in \mathbb Q[X]$ has a root $a \in \mathbb C$ of multiplicity $4$, then we have $m^4 |f$ in $\mathbb Q[X]$, where $m \in \mathbb Q[X]$ is the minimal polynomial of $a$. Then of course $\deg f=5$ yields $\deg m=1$, i.e. $a \in \mathbb Q$. Thus any root of multiplicity $4$ is actually a rational root, which (as the OP already figured out) would be either $1$ or $-1$ and you can just check that they are not (of course just by plugging in and not by doing the division). Thus the polynomial has no root of multiplicity $4$.
A: The derivative  of $P(x)=x^5-5x^4+1$ is $P'(x)=5x^3(x-4).$ Now $P'(x)>0$ for $x>4$ and also $P'(x)>0$ for $x<0.$  And $P'(x) <0$ for $x\in  (0,4).$ 
So $P$ is strictly increasing on  $(-\infty,0],$ strictly decreasing on $[0,4],$ and strictly increasing on $[4,\infty)$.
Therefore, from that, and from  $P(-1)<0<P(0)$ and $P(4)<0<P(5),$ we conclude that there are exactly 3 real solutions to $P(x)=0,$ one in $(-1,0),$ one in $(0,4),$ and one in $(4,5).$ The 2 other solutions to $P(x)=0$ must be  complex, of the form $a\pm bi,$ with real $a, b$ with $b\ne 0.$
A: Since $P(x)'=5x^3(x-4)$, the possibities are only two.
$P(x)=(x+a)(x-4)^4$ or $(x+b)x^4$
But these never become OP's.
A: extended gcd in $\mathbb Q[x].$ the original polynomial and its derivative are coprime; there are no multiple roots, not even double.
$$  \left(   x^{5}  - 5 x^{4}  + 1 \right)  $$ 
$$  \left(   x^{4}  - 4 x^{3}  \right)  $$ 
$$  \left(   x^{5}  - 5 x^{4}  + 1 \right)  =  \left(   x^{4}  - 4 x^{3}  \right)  \cdot \color{magenta}{ \left(   x  - 1 \right)}  +  \left(   - 4 x^{3}  + 1 \right)  $$ 
 $$  \left(   x^{4}  - 4 x^{3}  \right)  =  \left(   - 4 x^{3}  + 1 \right)  \cdot  \color{magenta}{ \left(   \frac{  -  x  + 4 }{ 4 }  \right)}  +  \left(   \frac{  x  - 4 }{ 4 }  \right)  $$ 
 $$  \left(   - 4 x^{3}  + 1 \right)  =  \left(   \frac{  x  - 4 }{ 4 }  \right)  \cdot  \color{magenta}{ \left(   - 16 x^{2}  - 64 x  - 256 \right)}  +  \left( -255  \right)  $$ 
 $$  \left(   \frac{  x  - 4 }{ 4 }  \right)  =  \left( -255  \right)  \cdot  \color{magenta}{ \left(   \frac{  -  x  + 4 }{ 1020 }  \right)}  +  \left( 0 \right)  $$ 
 $$ \frac{ 0}{1} $$ 
 $$ \frac{ 1}{0} $$ 
 $$  \color{magenta}{ \left(   x  - 1 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x  - 1 \right) }{ \left( 1  \right) } $$ 
 $$   \color{magenta}{\left(   \frac{  -  x  + 4 }{ 4 }  \right)}   \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  -  x^{2}  + 5 x  }{ 4 }  \right) }{ \left(   \frac{  -  x  + 4 }{ 4 }  \right) } $$ 
 $$   \color{magenta}{\left(   - 16 x^{2}  - 64 x  - 256 \right)}   \Longrightarrow  \Longrightarrow  \frac{  \left(  4 x^{4}  - 4 x^{3}  - 16 x^{2}  - 319 x  - 1 \right) }{ \left(  4 x^{3}  - 255 \right) } $$ 
 $$  \color{magenta}{ \left(   \frac{  -  x  + 4 }{ 1020 }  \right)}   \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  -  x^{5}  + 5 x^{4}  - 1 }{ 255 }  \right) }{ \left(   \frac{  -  x^{4}  + 4 x^{3}  }{ 255 }  \right) } $$ 
 $$  \left(   x^{5}  - 5 x^{4}  + 1 \right)  \left(   \frac{ 4 x^{3}  - 255 }{ 255 }  \right)  -  \left(   x^{4}  - 4 x^{3}  \right)  \left(   \frac{ 4 x^{4}  - 4 x^{3}  - 16 x^{2}  - 319 x  - 1 }{ 255 }  \right)  =  \left( -1  \right)  $$ 
A: Let $f(x) = x^5 + 5x^4 + 1.$
Rather than perform division of polynomials, a simple way to test possible rational roots is just to plug them into the equation. For $x=-1$ and $x=1$ this gives us
\begin{align}
f(-1) &= (-1)^5-5(-1)^4 + 1 = -5, \\
f(1)  &= (1)^5-5(1)^4 + 1 = -3.
\end{align}
But of course $f(0) = 1.$ So there is one root between $-1$ and $0$ and another between $0$ and $1.$ There must also be a root greater than $1$ since  we know $f(x)$ eventually grows without bound.
That's three distinct roots. If any of these had multiplicity $4,$
then $f(x)$ would have a total of at least six roots (including multiple roots).
But $f(x)$ is only fifth degree and can have at most five roots.
Therefore $f(x)$ has no roots of multiplicity $4.$
A: The "easy" part of the fundamental theorem of algebra says that a polynomial of degree $n$ has at most $n$ roots, counting multiplicities.  (The "hard" part says it has exactly $n$ roots, but we only need the "at most" part here.). Therefore, if $x^5-5x^4+1$ had a root of multiplicity $4$, it would have at most $2$ distinct roots, namely the the one of multiplicity $4$ and at most one other.  But the polynomial, which is of odd degree, is positive at $x=0$ and negative at $x=1$, so it has at least three distinct real roots (one less than $0$, one between $0$ and $1$, and one greater than $1$).  Thus it cannot have a root of multiplicity $4$.
Remark: This approach only rules out roots of multiplicity $4$ or $5$; it does not rule out the possibility of multiplicy $2$ or $3$.  But it avoids doing any explicit algebra beyond evaluating the polynomial in question at two points.
Additional remark:  This answer is essentially the same as David K's.  For some reason I didn't see his answer when I initially posted.
A: Note that for $n\geqslant 1$, a polynomial $f$ has a root of multiplicity $(n+1)$ if and only if $f$ shares a root with $f^{(n)}$. 
Consider $f = x^5 - 5x^4 + 1, n=3$. 
Then $f''' = 60x^2-60x=60x(x-1)$. 
The only roots of $f'''$ are $0,1$, which are not roots of $f$. 
Thus, $f$ and $f'''$ share no roots, and hence $f$ has no roots of multiplicity $4$. 
