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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  
\[f(x)\]
 , a contradiction.
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.

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