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Modulo three, the quintic has no linear factor: its value is $-1$ at every element of ${\mathbb{F}}_3$. The monic irreducible quadratics over this field are $t^2+1$, $t^2+t+2$ and $t^2+2t+2$. Division leaves respectively $-1$, $t-1$ and $t-1$, all nonzero. A reducible quintic has a factor of degree at most two, so it is irreducible modulo three and therefore over $\mathbb{Q}$. Its Galois group is transitive on five roots.

Modulo two its squarefree factorization isBoth factors are irreducible, so the Frobenius cycle type theorem gives an element that is a disjoint transposition times a three-cycle. Cubing it gives a transposition.

The prime-degree transposition criterion can be proved here: a transitive subgroup of prime degree containing a transposition is the whole symmetric group. Construct a graph on the five roots whose edges are the pairs swapped by all conjugates of this transposition. The group acts transitively on vertices and permutes connected components, so all component sizes are equal and divide five. An edge rules out size one; the graph is connected. Edge transpositions of any connected graph generate the full symmetric group, for instance by transporting swaps along a spanning tree. All belong to the Galois group, provingThe prime-degree condition is essential to the short transposition argument.