Prove that $x^5 - 5x + 1$ has no double roots in $\mathbb{C}[x]$

Question

I am asked this question:

Prove that $x^5 - 5x + 1$ has no double roots in $\mathbb C[x]$.

Now here's what I said:

$p(x) \in \mathbb C[x]$ has no double roots if and only if $gcd(p,p') = 1$. Now we need to prove that. So: $p(x) = x^5 - 5x + 1$ and $p'(x) = 5x^4 - 5$ and I start doing the Euclidean algorithm and I got to a point where I have to do $\frac{-4x+1}{ \frac{5}{4}x^3-5 }$, which we obviously can not do.

What do we do in such a situation? What does that mean that while doing the Euclidean algorithm we can not continue dividing?

Answer 1

You are almost finished. $\gcd(-4x+1, \frac{5}{4} x^3-5)=1$

Answer 2

In the Euclidean algorithm, you divide the larger by the smaller. A cubic polynomial is larger than a linear polynomial.

Computing gcds by finding the prime factorization is often a feasible alternative. In this setting, prime factorization means "factor", and is almost synonymous with finding its roots.