How to show $x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 = 0$ have five different roots?

Question

I have to prove this fact when determining the elementary divisors of a matrix. The numerical solution confirms that this equation indeed has five different roots (one real root, two pairs of conjugate complex roots), but how to show it theoretically? I looked into its derivatives but it didn't solve the problem.

Answer 1

To tell if $f$ has any multiple zeros, take the GCD of $f$ and $f'$. (Any multiple zero of $f$ is also a zero of $f'$.)

In this case $$ \gcd(f,f') = 1 , $$ so there are no multiple zeros, so there are $5$ simple zeros.

Euclidean algorithm computation. $f(x) = x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$, $f'(x) = 5x^4+20x^3+12x^2+6x+2$.

Divide $x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$ by $5x^4+20x^3+12x^2+6x+2$:
Remainder is $-{\frac {12\,{x}^{3}}{5}}-\frac35\,{x}^{2}+\frac25\,x+\frac35$.
Divide $5x^4+20x^3+12x^2+6x+2$ by $-{\frac {12\,{x}^{3}}{5}}-\frac35\,{x}^{2}+\frac25\,x+\frac35$. Remainder is ${\frac {391\,{x}^{2}}{48}}+{\frac {83\,x}{8}}+{\frac{107}{16}}$.
Divide $-{\frac {12\,{x}^{3}}{5}}-\frac35\,{x}^{2}+\frac25\,x+\frac35$ by ${\frac {391\,{x}^{2}}{48}}+{\frac {83\,x}{8}}+{\frac{107}{16}}$.
Remainder is $-{\frac {116000\,x}{152881}}-{\frac{216624}{152881}}$.
Divide ${\frac {391\,{x}^{2}}{48}}+{\frac {83\,x}{8}}+{\frac{107}{16}}$ by $-{\frac {116000\,x}{152881}}-{\frac{216624}{152881}}$.
Remainder is the constant $\frac{13220690237}{841000000}$.
Conclusion: $f$ has no multiple zero. If $a$ were a multiple zero, then $x-a$ divides both $f$ and $f'$, and therefore all the remainders in the calculation. But of course $x-a$ cannot divide a constant.

Answer 2

The discriminant of $x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$ is $86477$. Since it's not zero, all roots are simple.