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?
Best Answer
$$f(x)=x^4+x^3+x^2+x+1$$ if $x=1$ then $f(1)=5$ hense $f(1)>0$
if $x\neq 1$ then $f(x)=\frac{x^5-1}{x-1}$
$(x^5-1)$ and $(x-1)$ have the same sign any $x$. Hence $f(x)>0$