$t^4+t^3+t^2+t+1 $ has no linear factors

Question

I am trying to understand why the given polynomial has no linear factors over $\mathbb{Q}$. I am trying to do it using elementary methods (no eisenstein criterion etc to show it's irreducible).

I have read that it suffices that $t = \pm 1$ is not a root, why is that?

Answer 1

This is the rational root theorem, which says if a function has roots in $\mathbb Q$, then the denominator divides the first term, and the numerator divides the last term. Since the first term is one, any rational root is an integer dividing 1 - i.e. just $\pm 1$.

Now I think a little more has to be said, because you did not specify that you are only looking for roots in $\mathbb Q$. In this case, you don't know if there are irrational roots. To check this, you could use calculus, check to see that there are no minima with negative values. This would also assure you there are no roots.

Here is a link for the rational root theorem, with an elementary proof.

https://www.artofproblemsolving.com/wiki/index.php?title=Rational_Root_Theorem

Answer 2

It is easier to find the roots of $$p(t)=t^5-1 = (t-1)(t^4+t^3+t^2+t+1)\text{.}$$

If $p(t)= 0$, then $t^5 = 1$ and it follows that $|t^5|=|t|^5 = |1| = 1$, hence $|t| = 1$, hence the only two candidates are really just $-1$ and $1$.

Answer 3

Firstly, Since $$t^5-1=(t-1)(t^4+t^3+t^2+t+1)$$ will gives all $5$-th of unity except $1$ are roots (which are not real numbers too) of $$t^4+t^3+t^2+t+1.$$

Secondly, Since $t\ne 0$, if we divede $t^4+t^3+t^2+t+1$ by $t^2$ we will get $$t^2+t+1+\frac1t+\frac1{t^2}.$$ Say $u=t+\frac1t$ then $$u^2=t^2+\frac{1}{t^2}+2.$$ Hence we will get $$t^4+t^3+t^2+t+1=u^2+u-1$$ which is a quadratic equation. We can find the roots and then using $u=t+\frac1t$ we will solve another quadratic equations, which are simple.

Moreover, $t^4+t^3+t^2+t+1$ is $5$th cyclotomic polynomial and all cyclotomic polynomials are irreducible over $\mathbb Q$.

Furthermore, let $f(t)=t^4+t^3+t^2+t+1$ and use Eisenstein Criteria for $f(t+1)$ with the prime $5$ and conclude $f(t)$ is irreducible with knowledge of that $f(t+1)$ is irreducible.