3q^2 + 16q + 5
I look for a solution of the form (3q-a)(q-b). Expanding $\displaystyle 3q^2 - (3b+a)q +ab$. 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 $\displaystyle 3q^2 + 16q + 5= (3q+1)(q+5)$.
$\displaystyle P(q)=3q^2 + 16q + 5$
The rational roots theorem suggests that $\displaystyle \pm 5$, $\displaystyle \pm 1/3$ and $\displaystyle \pm 5/3$ are the only possible rational roots for this quadratic in $\displaystyle q$. Descartes rule of signs tells you it has no positive roots, so that reduces the list to $\displaystyle -5, -1/3$ and $\displaystyle -5/3$. Now trying these out shows that $\displaystyle P(-5)=0$ and $\displaystyle P(-1/3)=0$, so we have:
$\displaystyle P(q)=k(q+1/3)(q+5)$
and to match the leading coefficient of $\displaystyle P(q)$ we must have $\displaystyle k=3$ so:
$\displaystyle P(q)=(3q+1)(q+5)$
CB