Show $t^5-5t+1$ has no multiplicities

Question

Problem

Show that $f(t)=t^5-5t+1$ has no repeated roots.

My initial idea

I write $f$ in terms of linear factors $(t-\alpha_1)...(t-\alpha_5)$ where all the $\alpha_i$ make up the five roots of $f$. I’d like to show that each possibility of there even existing multiplicities in the factorization leads to a contradiction. The following statements are all that is necessary to prove the original question without loss of generality and after renumbering the $\alpha_i$. The first case would be that $f(t)=(t-\alpha_1)^2(t-\alpha_2)(t-\alpha_3)(t-\alpha_4)$. The second would be that $f(t)=(t-\alpha_1)^2(t-\alpha_2)^2(t-\alpha_3).$ The third case would be that $f(t)=(t-\alpha_1)^3(t-\alpha_2)(t-\alpha_3)$, and so on proving the rest of these possibilities until they are all exhausted out.

My question

Obviously, this is time consuming and there’s likely a better solution. I would like any suggestions on how to solve the problem quicker without this exhaustive method.

Any thoughts would be appreciated.

Remark

I tagged irreducible polynomials and gcd/lcm because I am not sure if any suggestions would have to do with them, yet I still suspect that they would.

Answer 1

$f(t)=t^5-5t+1 \implies f'(t)=5t^4-5, f'(1)=0,f'(-1)=0, f'(i)=0, f'(-i)=0$ but $f(1), f(\pm 1), f(\pm i)$ are non-zero. So there exist not even a root with multiplicity 2.

Edit: For multiplicity of orfer $m$ at $x=a$ we must have $f(a)=f'(a)=...f^{(m-1)}(a)=0$. $k(x)=(x-1)^4$ has multiplicity of order 4 at $x=1$.

Answer 2

You can solve this problem if you use discriminant of polynomial P = resultant of P and P' (its derivative). You may look at definition of resultant there - https://en.wikipedia.org/wiki/Resultant

$$ \phi (A, B) = \begin{vmatrix} a_0 & 0 & \cdots & 0 & b_0 & 0 & \cdots & 0 \\ a_1 & a_0 & \cdots & 0 & b_1 & b_0 & \cdots & 0 \\ a_2 & a_1 & \ddots & 0 & b_2 & b_1 & \ddots & 0 \\ \vdots &\vdots & \ddots & a_0 & \vdots &\vdots & \ddots & b_0 \\ a_d & a_{d-1} & \cdots & \vdots & b_e & b_{e-1} & \cdots & \vdots\\ 0 & a_d & \ddots & \vdots & 0 & b_e & \ddots & \vdots \\ \vdots & \vdots & \ddots & a_{d-1} & \vdots & \vdots & \ddots & b_{e-1} \\ 0 & 0 & \cdots & a_d & 0 & 0 & \cdots & b_e \end{vmatrix},$$

where $A$ and $B$ - polynomials with coefficients $a_i$ and $b_i$.

In our case:

$$\phi (P, P') = \begin{vmatrix} 1&0&0&0& 5&0&0&0&0\\ 0&1&0&0& 0&5&0&0&0\\ 0&0&1&0& 0&0&5&0&0\\ 0&0&0&1& 0&0&0&5&0\\ -5&0&0&0& -5&0&0&0&5\\ 1&-5&0&0& 0&-5&0&0&0\\ 0&1&-5&0& 0&0&-5&0&0\\ 0&0&1&-5& 0&0&0&-5&0\\ 0&0&0&1& 0&0&0&0&-5\\ \end{vmatrix} = -796875.$$

This determinant is not zero iff there are no repeated roots. It is easy to calculate in wolfram mathematics.