Doubt in IMO $1993$ Problem 1

Question

Question -

Given an integer $n>1$, consider the polynomial $f(x)=x^{n}+5 x^{n-1}+3 .$ Prove that there are no nonconstant polynomials $g(x), h(x)$ with integer coefficients such that $f(x)=g(x) h(x)$

Solution -

By the (extended) Eisenstein criterion, $f$ has an irreducible factor of degree at least $n-1$, since $f$ has no integer zeros, it must be irreducible. $\triangle$

now i did not understand there last line i.e since $f$ has no integer zeros, it must be irreducible

I understand that using (extended) Eisenstein criterion (taking $p=3$ and $k=n-2$) they proved that $f$ has an irreducible factor of degree at least $n-1$,

I also proved that f has no integer roots by using rational root theorem but i did not see how that proves that f is irreducible ?

I know that if f has no integer roots then it is irreducible in Z[X] but this is valid for degree only $2$ or $3$ not for all n .

Answer 1

Supposedly you're reading here? Check out Extended Eisenstein's Criterion, as written:

Let $P(x) = a_nx^n + \cdots + a_1 x + a_0$ be a polynomial with integer coefficients. If there exists a prime $p$ and integer $k \in \{0, 1, \ldots, n - 1\}$ such that $$p \mid a_0, \ldots, a_k, \; p\not\mid a_{k+1}, \text{ and } p^2 \not\mid a_0,$$ then $P(x)$ has an irreducible factor of degree greater than $k$.

In particular, if $p$ can be taken so that $k = n - 1$, then $P(x)$ is irreducible.

So, in the case of $f(x) = x^n + 5x^{n-1} + 3$, we can take $p = 3$ and $k = n - 2$ (you should verify the hypotheses hold with these numbers). Extended Eisenstein's Criterion implies that there is an irreducible factor $g(x)$ of degree at least $n - 1$. If it's degree $n$, then we are done, and $f(x)$ must be irreducible. Otherwise it's degree $n - 1$, in which case we can write $$f(x) = g(x)h(x)$$ for some degree $1$ polynomial with integer coefficients. Such a polynomial has a rational root, and hence so must $f$. But, as you said, $f$ has no rational root, $f$ must be irreducible.