Let $f(x) = a_0\!+\!a_1x\!+\cdots+\!a_nx^n$ , 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 $ab = 1$ .