+145 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 \(x\) except -5, 0, and 4. Find \(H/G\).
+129899 We can use partial fractions, here
Note that x^3 + x^2 -20x can be factored as ( x^2 - 4x) ( x + 5)
x^2 - 2x + 10 G H
______________ = _____ + ________
( x^2 - 4x)(x + 5) x + 5 x^2 - 4x
Multiply through by ( x + 5) ( x^2 - 4x)
x^2 - 2x + 10 = G(x^2 - 4x) + H (x + 5) simplify
x^2 - 2x + 10 = Gx^2 - 4Gx + Hx + 5H equate coefficients
1 = G
-2 = H - 4G
10 = 5H
It's obvious that G = 1 and H = 2
So
H / G = 2 / 1 = 2
