How to show that x^2−2 is irreducible polynomial over Q?

Problem : show that x^2−2 is irreducible polynomial over Q.

Answer :

Given integral domain (Q , +, . ) where Q is the set of all rational numbers.

Given a non zero polynomial

\[ f(x) =x^2−2 \]

Then f(x) will be reducible polynomial in (Q[x] , + , . ) if there exists non constant polynomials g(x) and h(x) in Q[x] such that f(x)= g(x). h(x) where degree g(x) is less than degree of f(x) . Otherwise , f(x) will be irreducible polynomial.

Since only factors of \[ f(x) =x^2−2 \] are \[(x-\sqrt{2})(x+\sqrt{2})\]

and both \[g(x)=(x-\sqrt{2}), h(x) =(x+\sqrt{2})\] are not polynomial in (Q[x] , + , . ).

Because \[\sqrt{2}\] is not rational number. It is an irrational number.

Therefore x^2−2 is irreducible polynomial over Q.