Use the <Eisenstein criterion> with prime $5$. The leading coefficient is one; every lower coefficient, including the zero coefficients, is divisible by $5$; and the constant coefficient $5$ is not divisible by $25$. The polynomial is monic and hence primitive. Therefore \b[it is irreducible over $\mathbb Q$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 2F / i / Solution

 Solution (source code)