SOLUTION: {{{(3x)/(2x+5)-(5x+2)/(2x-5)=(Ax^2+Bx+C)/((2x+5)(2x-5))}}} Find the smallest values of the numbers |A|, |B|, and |C|

Algebra ->  Absolute-value -> SOLUTION: {{{(3x)/(2x+5)-(5x+2)/(2x-5)=(Ax^2+Bx+C)/((2x+5)(2x-5))}}} Find the smallest values of the numbers |A|, |B|, and |C|      Log On


   

Question 716356:
Find the smallest values of the numbers |A|, |B|, and |C|
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!

put the left side over a single denominator
%283x%282x-5%29-%282x%2B5%29%285x%2B2%29%29%2F%28%282x%2B5%29%282x-5%29%29=%28Ax%5E2%2BBx%2BC%29%2F%28%282x%2B5%29%282x-5%29%29
If the denominators are equal, the numerators are equal, so we can write
3x(2x-5) - (2x+5)(5x+2) = Ax^2 + Bx + C
6x^2 - 15x - (10x^2 + 4x + 25x + 10) = Ax^2 + Bx + C
6x^2 - 15x - 10x^2 - 29x - 10 = Ax^2 + Bx + C
Combine like terms
6x^2 - 10x^2 - 15x - 29x - 10 = Ax^2 + Bx + C
-4x^2 - 44x - 10 = Ax^2 + Bx + C
Absolute values
A = 4
B = 44
C = 10