The integers $G$ and $H$ are chosen such that
\[\frac{G}{x+5}+\frac{H}{x^2-4x}=\frac{x^2-2x+10}{x^3+x^2-20x}\]
for all real values of except $-5$, $0$, and $4$. Find $H/G$.
cross multiply the left side
gx^2 -g 4x + hx + 5 h = x^2 -2x+10 so g = 1 h = 2 take it from here !