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?
 A: $$f(x)=x^4+x^3+x^2+x+1$$
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 $f(x)>0$
A: There is a theorem about positive polynomials being expressible as the sum of squares. Here $$f(x)=\left(x^2+\frac x2\right)^2 +\frac 34\left(x+\frac 23\right)^2+\frac 23$$
obtained by successively completing the square. Personally I prefer some of the simpler methods. For example one can observe that the roots of $f(x)$ are the complex fifth roots of $1$, so there are no real zeros.
A: 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
$$t=\dfrac1{1-t+t^2-t^3}>1,\quad\text{contradiction}.$$
A: 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.
