I look for a solution of the form (3q-a)(q-b). Expanding . Equating the coefficients of like terms gives ab=5 and 3b+a=-16. Thinking for a moment will convince you that b=-5 and a=-1 satisfy these equations. So .
The rational roots theorem suggests that , and are the only possible rational roots for this quadratic in . Descartes rule of signs tells you it has no positive roots, so that reduces the list to and . Now trying these out shows that and , so we have:
and to match the leading coefficient of we must have so: