Let  $f(x)=a_{0}\!+\!a_{1}x\!+\cdots+\!a_{n}x^{n}$  be a polynomial with complex coefficients $a_{{\nu}}$ and with the degree  $n>0$.  If $f(x)$ can not be written as product of two polynomials with positive degrees and with coefficients in the field  $\mathbb{Q}(a_{0},\,a_{1},\,\ldots,\,a_{n})$,  then the polynomial $f(x)$ is said to be irreducible.  Otherwise, $f(x)$ is reducible.

Examples.  All linear polynomials are irreducible.  The polynomials $x^{2}\!-\!3$, $x^{2}\!+\!1$ and $x^{2}\!-\!i$ are irreducible (although they split in linear factors in the fields $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(i)$ and $\mathbb{Q}(\frac{1\!+\!i}{\sqrt{2}})$, respectively).  The polynomials $x^{4}\!+\!4$ and $x^{6}\!+\!1$ are not irreducible.

The above definition of irreducible polynomial is special case of the more general setting where $f(x)$ is a non-constant polynomial in the polynomial ring $K[x]$ of a field $K$; if $f(x)$ is not expressible as product of two polynomials with positive degrees in the ring $K[x]$, then $f(x)$ is irreducible (in $K[x]$).

Example.  If $K$ is the Galois field with two elements (0 and 1), then the trinomial $x^{2}\!+\!x\!+\!1$ of $K[x]$ is irreducible (because an equation  $x^{2}\!+\!x\!+\!1=(x\!+\!a)(x\!+\!b)$  would imply the two conflicting conditions  $a\!+\!b=1$  and  $ab=1$).