<PRE><B>Question 1030202</B>
Solve the partial fraction decomposition of the rational expression: 
(3x^2+49) / (x(x+7)^2) = (A/x) + (B/(x+7)) + (C/(x+7)^2)
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Equate the numerators of the two sides::
3x^2 + 49 = A(x+7)^2 + B(x(x+7)) + Cx
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3x^2 + 49 = A(x^2+14x + 49) + B(x^2+7x) + Cx
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3x^2 + 49 = (A+B)x^2 + (14A +7B +C)x + 49A 
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Equate the proper coefficients:
A+B = 3
14A + 7B + C = 0
49A = 49
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Solve for A, B, and C
A = 1
B = 3-1 = 2
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C = -14-14 = -28
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(3x^2+49) / (x(x+7)^2) = (A/x) + (B/(x+7)) + (C/(x+7)^2)
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(3x^2+49) / (x(x+7)^2) =  = 1/x + 2/(x+7) - 28/(x+7)^2
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Cheers,
Stan H.
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3x^2 + 49 = 1/(x+7)^2 + 2/</PRE>