jmc

Setting $x=3$ and $x=-2,$ we get the two equations $$\begin{array}{rl}27a+9b+3c+d& =0,\\ -8a+4b-2c+d& =0.\end{array}$$Subtracting these two equations eliminates $d$ and gives $$35a+5b+5c=0.$$Thus, $b+c=-7a,$ so $\frac{b+c}{a}=\boxed{-7}.$