Is it true that the polynomial $x^5 + x + 1$ has all its simple roots?

Question

I know that $a$ is root of $p$ iff $p(a)=p'(a)=0$

$p(x)= x^5 + x + 1$

$p'(x)= 5x^4 + 1$ then $p'(a)=0$ iff $5a^4 + 1 = 0$ iff $a^4=-1/5$

Now, $p(a)= a^5 + a + 1 = a. a^4 + a + 1 = (-1/5) a + a + 1 = (4/5) a + 1 = 0$

then, $a = 1 - 5/4$ so $a = -1/4$. But $(-1/4)^4 ≠ -1/5$ so there is no solution. So $p$ does not have multiple roots.

So $p$ has all its simple roots.

Is my exercise resolution correct? Are the steps well explained and justified?

Answer 1

You made a minor mistake when you wrote that $5a^4+1=0\iff a=-1/5$. You should have written $a^4=-1/5$ instead.

The rest is correct, assuming that you are working over a field with characteristic $0$.

Answer 2

Consider $f(x)=x^5+x+1$ and $f'(x)=5x^4+1$ like you said. Since $f'(x)$ is always positive, we can conclude that $f(x)$ is always increasing. This means that the function only crosses the x-axis once (since $f(x)$ is of odd degree). Therefore we can conclude that $f(x)$ has only one real root, and since the function is always increasing, it must be a simple root.