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