Let K, the field generated by the coefficients of f, have t=2^n elements. If p=q=0, f has a triple root. If q is not 0, then f is separable. Assume q not 0. If p=0 and n odd, then f has exactly one root in K. If p=0 and n even then f has 0 or 3 roots in K. Assume now that f has at least 1 root in K. Let tr be the trace of K over the prime field. Put B=1+(p^3/q^2). Then f has 3 roots iff tr(B)=0.

The question if f has no roots (i.e. f irreducible) can be done by Berlekamp's algorithm.

Let $K$, the field generated by the coefficients of $f$, have $t=2^n$ elements. If $p=q=0$, $f$ has a triple root. If $q$ is not $0$, then $f$ is separable. Assume $q$ not $0$. If $p=0$ and n odd, then $f$ has exactly one root in $K$. If $p=0$ and $n$ even then $f$ has $0$ or $3$ roots in $K$. Assume now that $f$ has at least 1 root in $K$. Let $tr$ be the trace of $K$ over the prime field. Put $B=1+\frac{p^3}{q^2}$. Then $f$ has $3$ roots iff $tr(B)=0$.

The question if $f$ has no roots (i.e. $f$ irreducible) can be done by Berlekamp's algorithm.