Let $F$ be a unique factorization domain and $f \in F$. Show that the radical of the ideal generated by $f$ is principal, i.e. $\sqrt{(f)} = (g)$.
Since $F$ is a unique factorization domain we can write an unique decomposition of $f$, i.e. write $f = \prod_{i} p_i^{n_i}$. If we choose $h \in \sqrt{(f)}$, then by definition of the radical, there exists $m \in N^*$ such that $h^m \in (f)$. Thus we can write $h^m = k f = k \prod_{i} p_i^{n_i}$. Since $h \in F$, we can again decompose $h$ into irreducible elements, i.e. $h = \prod_{j} s_j$.
Thus we have, $\prod_{j} s_j^m = k \prod_{i} p_i^{n_i}$.
How to proceed now?
Note that you have shown that $p_i \mid h^m$ for all $i$ and hence, $p_i \mid h$ for all $i$. (Why?)
Using that, you can see that $h \in \left(\prod_ip_i\right).$ Thus, $\sqrt{(f)} \subset \left(\prod_ip_i\right)$. Can you show the other inclusion as well?
To continue your approach, you could further decompose each $s_j$ as a product of primes and conclude that each $p_i$ divides some $s_j$ and deduce the same as above.