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We can factor the numerator and denominator to get $$\frac{x^2+3x+2}{x^3+x^2-2x}=\frac{(x+1)(x+2)}{x(x-1)(x+2)}.$$In this representation we can immediately see that there is a hole at $x=-2$, and vertical asymptotes at $x=1$ and $x=0$. There are no more holes or vertical asymptotes, so $a=1$ and $b=2$. If we cancel out the common factors we have $$\frac{(x+1)(x+2)}{x(x-1)(x+2)}=\frac{x+1}{x^2-x}.$$We can now see that as $x$ becomes very large, the $x^2$ term in the denominator dominates and the graph tends towards $0$, giving us a horizontal asymptote. Since the graph cannot have more than one horizontal asymptote, or a horizontal asymptote and a slant asymptote, we have that $c=1$ and $d=0$. Therefore, $a+2b+3c+4d=1+2\cdot 2+3+0=\boxed{8}.$