Prove that every zero of $z^{4}-5z+1$ has multiplicity 1 Let $z \in \mathbb{C}$ I need to prove $z^{4}-5z+1$ has only zeros with multiplicity one "with minimal calculation". I've tried using the argument principle, but that requires me to find a curve that surrounds every zero. The only other thing I can think of is proving that $z^{4}-5z+1$ can only be a product of degree 1 irreducible polynomials but that doesn't seem like "minimal calculation". Are there any quick ways of solving this?
 A: A double (or worse) root would be a root of the derivative, too, but 
$$ 4f(z)-z f'(z) = 4-15 z $$
only vanishes at $z=\frac{4}{15}$, which is not a root for $f'(z)$. It follows that $f(z)$ and $f'(z)$ do not share any root, i.e. that all the roots of $f(z)$ are simple.

The idea above does not come out of the blue: it actually is the first step in the computation of a Sturm chain.
A: Note that
$$f'(z)=4z^3-5=0\implies z=\left(\frac54\right)^\frac13\implies z_1=\left(\frac54\right)^\frac13,z_2=\left(\frac54\right)^\frac13e^{\frac23\pi i},z_3=\left(\frac54\right)^\frac13e^{\frac43\pi i}$$
and 
$$z_1^{4}-5z_1+1=-3z_1^4+1=-3\left(\frac54\right)^\frac43+1\neq 0$$
$$z_2^{4}-5z_2+1=-3z_2^4+1=-3\left(\frac54\right)^\frac43e^{\frac83\pi i}+1\neq 0$$
$$z_3^{4}-5z_3+1=-3z_3^4+1=-3\left(\frac54\right)^\frac43e^{\frac{16}3\pi i}+1\neq 0$$
A: Alt. hint: the polynomial has no negative real roots, and $0$ or $2$ positive ones by the rule of signs. Since $p(0) \gt 0$ and $p(1) \lt 0$ it has in fact two positive roots, one less than $1$ and the other greater than $1$, thus distinct. The remaining two (non-real) complex roots must also be distinct (why?).
A: Assume $a\in \mathbb C$ is (at least) a double root of the given polynomial $f(z) = z^4-5z+1$. Then $a$ is also a root of $f'(z) = 4z^3-5$. We need to get a quick contradiction. From
$$
0=\Big(a^4-5a+1\Big)-a\cdot\frac 14\Big(4a^3-5\Big)=-\frac {15}4a+1
$$
we get $a=4/15$, which is not a root (of $f'$).
Note:
Using a computational aid, sage:
sage: R.<z> = PolynomialRing(QQ)
sage: f = z^4 -5*z +1
sage: # f has multiple roots iff its discriminant vanishes, but...
sage: f.discriminant()
-16619
sage: # The roots are approximatively...
sage: f.roots(QQbar, multiplicities=False)
[0.2003220662079570?,
 1.637305775417577?,
 -0.9188139208127669? - 1.484812643834945?*I,
 -0.9188139208127669? + 1.484812643834945?*I]

(I had to insert something new, since too many answers and comments were giving the same computation, so all humans think humanly, all computers computationally...) 
