Prove a polynomial has all roots different I need to prove that $P(x)=x^4+\zeta x+1$ where $\zeta\in\mathbb{R}$ and $\zeta\neq0$ has four different roots.
I have tried with the rule of signs of Decartes but it does not give enough information.
Any ideas?
 A: The statement is false. Set
$$\zeta=-\frac{4}{\sqrt[4]{27}}$$
Then the number $1/\sqrt[4]{3}$ is a repeated root for $P(x)$ as you can check directly.
How did I find it?
Divide $P(x)$ by $x-r$, assuming that $r$ is a root of $P$. The quotient is
$$Q(x)=x^3 + rx^2 + r^2x + (\zeta+r^3)$$
and $r$ is a root of $Q(x)$ if and only if $4r^3+\zeta=0$. Now compute $P(r)$ with this new information and find that $r=1/\sqrt[4]{3}$.

Let's do it in a different way. Consider $P$ as a function and set, for convenience, $\zeta=-4\lambda^3$ (that has only one real solution $\lambda$; its derivative is
$$P'(x)=4x^3 - 4\lambda^3$$
which has a unique root in the real numbers, namely $\lambda$. Therefore the polynomial has at most two real roots.
The absolute minimum of $P$ is attained at $\lambda$ and its value is
$$P(\lambda)=\lambda^4 - 4\lambda^4 + 1 = 1 - 3\lambda^4.$$


*

*If $1-3\lambda^4>0$, the polynomial has no real roots.

*If $1-3\lambda^4=0$, the polynomial has a double root.

*If $1-3\lambda^4<0$, the polynomial has two distinct real roots.


Notice that $\lambda=1/\sqrt[4]{3}$ gives exactly $\zeta=-4/\sqrt[4]{27}$.
A: A number $x_0$ is a repeated root of $P$ precisely when it is also a root of its derivative. The derivative of $P$ is $P'(x) = 4x^3 + \zeta$. The roots of $P'$ are easy to find, and it's equally easy to check that they are not roots of $P$.
A: Your approach could be generalized to use Sturm's theorem. As the comment suggests, a naive direct attack on the problem should be effective too.
There are two standard, general methods related to repeated factors of a polynomial, shown below. You may have to do a little bit more work in both cases to deal with the possibility that the repeated factor may or may not have any real roots. Or maybe a little bit less work if you're lucky.
Test if $\gcd(P(x), P'(x)) = 1$
This will tell you if $P(x)$ has any repeated factors; in fact, it even gives some information about what the repeated factor is.
It's not hard to see why this test works: consider $f(x) = g(x)^2 h(x)$, then compute $f'(x)$ and factor it. (it's worth also considering $f(x) = g(x)^3 h(x)$)
The Euclidean algorithm is a general method for computing this $\gcd$. But in your case, there is an alternate approach available because $P'(x)$ is easy to factor.
Test if $\mathrm{Res}(P(x), P'(x)) = 0$
This only makes sense if you're familiar with resultants. A lot of useful information can be gleaned from resultant calculations -- e.g. in this case it would give you a polynomial in $\lambda$, and the roots of this polynomial are precisely the values of $\lambda$ for which $P(x)$ has a repeated factor.
(Note that there are a value of $\lambda$ for which $P(x)$ has a repeated factor, and some of them are even real! But the repeated factor doesn't have any real roots)
The polynomial computed by the resultant formula above is also called the discriminant of $P(x)$.
