Originally Posted by

**skeeter** $\displaystyle \frac{5x^2+7x-9}{4-8x + x^2-2x^3}$

$\displaystyle \frac{5x^2+7x-9}{4(1-2x) + x^2(1-2x)}$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)}$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{A}{1-2x} + \frac{Bx+C}{x^2+4}$

$\displaystyle 5x^2+7x-9 = A(x^2+4) + (Bx+C)(1-2x)$

$\displaystyle 5x^2+7x-9 = Ax^2+4A + Bx - 2Bx^2 +C - 2Cx$

$\displaystyle 5x^2+7x-9 = (A-2B)x^2 + (B-2C)x + (4A+C)$

equating coefficients ...

$\displaystyle A-2B = 5$

$\displaystyle B-2C = 7$

$\displaystyle 4A+C = -9$

solving the system of equations ...

$\displaystyle A = -1 \, , \, B = -3 \, , \, C = -5$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{1}{2x-1} - \frac{3x+5}{x^2+4}$

I agree with this solution.