We are given the equation $x^3 + 2x^2y - x^2 + 2xy + 4y^2 - 2y = 44$. The problem asks us to factor the left side of the equation with respect to $y$, then analyze the difference of the factors.

1. Problem Description

We are given the equation $x^3 + 2x^2y - x^2 + 2xy + 4y^2 - 2y = 44$. The problem asks us to factor the left side of the equation with respect to $y$, then analyze the difference of the factors.

2. Solution Steps

(1) Rearranging the left side of the equation in terms of $y$, we have

$2x^2y + 2xy + 4y^2 - 2y + x^3 - x^2 = 44$.

$4y^2 + (2x^2 + 2x - 2)y + (x^3 - x^2 - 44) = 0$.

However, the problem statement hints to factor it as $(x^A + By)(x + Cy - D)$.

Let's regroup the original expression:

$x^3 - x^2 + 2x^2y + 2xy + 4y^2 - 2y = 44$

$x^2(x - 1) + 2y(x^2 + x + 2y - 1) = 44$

Consider $(x^2+2y)(x-1+2y)$. Expanding this gives

$x^3 - x^2 + 2x^2y + 2xy + 4y^2 - 2y$.

Thus we have

$(x^2+2y)(x-1+2y)=44$

$(x^2 + 2y)(x + 2y - 1) = 44$

Comparing this to $(x^A + By)(x + Cy - D)$, we have $A=2, B=2, C=2, D=1$.

(2) $x^2 + 2y$ and $x + 2y - 1$ are integers, and their product is

4

4. We compute their difference.

$(x^2 + 2y) - (x + 2y - 1) = x^2 - x + 1 = x(x-1) + 1$

$x(x-1) + 1 = x(x-E) + F$

$x(x-1) + 1 = x^2 - x + 1$

Comparing the coefficients, we can see $E = 1$ and $F = 1$.

Thus, we have $(x^2 + 2y) - (x + 2y - 1) = x(x-1) + 1 = d$

$d = x^2 - x + 1$. Since $x^2 - x = x(x-1)$ is the product of two consecutive integers, it is an even number. Therefore, $x^2 - x + 1$ must be odd. So $d$ is odd. Also, $x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$, which is always positive.

Therefore, $d>0$ and $d$ is odd.

G corresponds to d>0, so G is

0. H corresponds to d being odd, so H is

4.

3. Final Answer

$A=2, B=2, C=2, D=1$

$E=1, F=1$

$G=0, H=4$