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?

Another proof by contradiction.
Suppose $x^4+x^3+x^2+x+1$ has a real root. Necessarily, $x<0$. Set $x=-t,\enspace t>0$. Rewrite the equation as
$$t^4-t^3+t^2-t=t(t-1)(t^2+1)=-1$$
It shows $t-1<0$, hence $\;0<t<1$.
On another hand, $\;t=\dfrac1{(1-t)(1+t^2)}=\dfrac1{1-t+t^2-t^3}$, and since $0<t<1$, we have $0<1-t+t^2-t^3<1-t+t^2=1-t(1-t)<1$, hence
$$t=\dfrac1{1-t+t^2-t^3}>1,\quad\text{contradiction}.$$
