A polynomial of degree 1 (also called a linear polynomial) is irreducible. The main theorem of algebra tells us that every polynomial in C[X] is a product of linear factors. So, the only complex irreducible polynomials are the linear ones.

In Q[X] there exist irreducible factors of larger degree. Put f(X) = X² + bX + c with b, c Q. Then f is reducible if there exists a q Q such that X - q | f(X). This is the same as f(q) = 0 (see exercise from the previous section). It is well known that f has a zero if and only if the discriminant of f is the square of a rational number. We conclude that f Q[X] is irreducible if and only if b² - 4c is not a square in Q.

Take b = 0 and c = -2, so that f(X) = X² - 2. Then f is irreducible in Q[X] according to the above-discussed criterion, but reducible in R[X].

Put R = Z/2Z. We prove that the polynomial f(X) = X⁴ + X + 1 is irreducible in R[X]. Since f(0) = f(1) = 1, there are no zeros, and for that reason no linear factors of f. Now suppose that f = g ·h with deg(g) = deg(h) = 2. Write g(X) = X² + aX + b and h(X) = X² + cX + d with a, b, c, d Z/2Z. Then f = gh implies the equations:

a = c,

ac = b + d,

ad + bc = 1,

bd = 1.

The last equation implies b = d = 1, so that the first and the third equations imply 0 = a + c = 1, which is a contradiction. It follows that f can in no way be written as a product of polynomials of lower degree, in other words, f is irreducible.

As for integers (compare with the example on the factorization record), it is not difficult to verify a factorization. However, it is not always easy to check whether the factors found are irreducible. A proof that a polynomial f Z[X] is irreducible can often be given by computing modulo p for a prime number p. This doesn't always work: there are polynomials f Z[X] which are irreducible but factorize modulo each prime p. An example is f(X) = X⁴ + 1. Modulo 2 it factors as (X + 1)⁴ and modulo 3 as (X² - X - 1)(X² + X - 1). It would be going too far to show that X⁴ + 1 is reducible modulo every prime.