## Eisenstein’s Irreducibility Criterion

Theorem
Let
$f(x) = a_0 + a_1 x + ... + a_n x^n$
be a polynomial with integer coefficients. Suppose a prime
$p$
divides each of
$a_0, a_1, ..., a_{n-1}$
(every coefficient except coefficient of
$x^n$
), and that
$p^2$
does not divide
$a_0$
. Then
$f(x)$
has no factors with integer coefficients.
Proof
Suppose
$f = g h$
for polynomials
$g, h$
with integer coefficients. Look at this factorization modulo
$p$
: we get
$f(x) = a_n x^n$
, so
$g(x) = b_d x^d$
,
$h(x) = c_e x^e$
for some constants
$b_d, c_e$
and for some integers
$d, e$
with
$d e = n$
. This implies the constant term of
$g(x)$
is a multiple of
$p$
, and similarly for the constant term of
$h(x)$
, hence
$p^2$
divides the constant term of
$f(x)$
Example
Take
$f(x)=x^3+4x^2+10x+14$

2 divides every coefficient except the coefficient of
$x^3$
and
$2^2=4$
does not divide the constant term 14. Therefore
$f(x)$
does not factorise.