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 $H/G$.
First, we factor the denominators, to get \[\frac{G}{x + 5} + \frac{H}{x(x - 4)} = \frac{x^2 - 2x + 10}{x(x + 5)(x - 4)}.\]We then multiply both sides by $x(x + 5)(x - 4)$, to get \[Gx(x - 4) + H(x + 5) = x^2 - 2x + 10.\]We can solve for $G$ and $H$ by substituting suitable values of $x$.  For example, setting $x = -5$, we get $45G = 45$, so $G = 1$.  Setting $x = 0$, we get $5H = 10$, so $H = 2$.  (This may not seem legitimate, because we are told that the given equation holds for all $x$ except $-5$, 0, and 4.  This tells us that the equation $Gx(x - 4) + H(x + 5) = x^2 - 2x + 10$ holds for all $x$, except possibly $-5$, 0, and 4.  However, both sides of this equation are polynomials, and if two polynomials are equal for an infinite number of values of $x$, then the two polynomials are equal for all values of $x$.  Hence, we can substitute any value we wish to into this equation.)

Therefore, $H/G = 2/1 = \boxed{2}$.
2
Level 5
Algebra
Problem: 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 $H/G$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the ratio \( H/G \), we start by simplifying the given equation. The left-hand side consists of two fractions, which we need to combine over a common denominator. The denominators are \( x + 5 \) and \( x^2 - 4x \). The expression \( x^2 - 4x \) can be factored as \( x(x - 4) \). Thus, the common denominator for the left-hand side is \( (x + 5)(x)(x - 4) \).

Next, we rewrite the left-hand side:

\[
\frac{G}{x + 5} + \frac{H}{x(x - 4)} = \frac{G \cdot x(x - 4) + H \cdot (x + 5)}{(x + 5)(x)(x - 4)}
\]

Now, we equate this to the right-hand side, which is:

\[
\frac{x^2 - 2x + 10}{x^3 + x^2 - 20x}
\]

We need to factor the denominator on the right-hand side. The expression \( x^3 + x^2 - 20x \) can be factored as \( x(x^2 + x - 20) \). The quadratic \( x^2 + x - 20 \) factors further into \( (x - 4)(x + 5) \). Therefore, the complete factorization of the denominator is:

\[
x(x - 4)(x + 5)
\]

Now, we can equate the numerators:

\[
G \cdot x(x - 4) + H \cdot (x + 5) = x^2 - 2x + 10
\]

Expanding the left-hand side gives:

\[
Gx^2 - 4Gx + Hx + 5H = (G)x^2 + (-4G + H)x + 5H
\]

Setting the coefficients equal to those on the right-hand side, we have the following system of equations:

1. \( G = 1 \) (coefficient of \( x^2 \))
2. \( -4G + H = -2 \) (coefficient of \( x \))
3. \( 5H = 10 \) (constant term)

From the third equation, we solve for \( H \):

\[
5H = 10 \implies H = 2
\]

Substituting \( H = 2 \) into the second equation:

\[
-4G + 2 = -2 \implies -4G = -4 \implies G = 1
\]

Now we have \( G = 1 \) and \( H = 2 \). To find the ratio \( H/G \):

\[
\frac{H}{G} = \frac{2}{1} = 2
\]

Thus, the final answer is \( 2 \).<|END_OF_LATENT|> The final answer is $2$.