Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots.
I can see two ways of proving it.
The first one is to notice, that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. It has the only root 1. And it is not the root of $x^4 + x^3 + x^2 + x + 1$. So, $x^4 + x^3 + x^2 + x + 1$ does not have roots.
Another way is to solve it as a palindromic polynomial. It does not have roots.
But is there any way to directly manipulate the expression to show that it is always greater than zero?

Assuming that exists $a
 $ such that $$a^{4}+a^{3}+a^{2}+a+1=0.
 $$ Now it's obvious that $a\neq0
 $. So we have $$0=\frac{\left(a^{4}+a^{3}+a^{2}+a+1\right)}{a}=a^{3}+a^{2}+a+1+\frac{1}{a}
 $$ so $$a^{3}+a^{2}+a+1+\frac{1}{a}=a^{4}+a^{3}+a^{2}+a+1\Longrightarrow\frac{1}{a}=a^{4}\Rightarrow1=a^{5}
 $$ so we have $$a=1
 $$ and this is absurd. Now since $f\left(x\right)=x^{4}+x^{3}+x^{2}+x+1
 $ is a continuous function, to see if it is positive it is sufficient see what happens at one point.
