\(\frac{s-1}{s(s-2)^2}\)
How can I expand this fraction?
\(\frac{A}{s} + \frac{B}{(s-2)} + \frac{C}{(s-2)^2}\)
right?
This gives me the equation
\(As^3 - 6As^2 + 12As - 8A Bs^3 - 4Bs^2 + 4Bs + Cs^2 - 2Cs = s-1\)
so that
(1) A + B =0
(2)- 6A - 4B + C = 0
(3) 12A + 4B - 2C = 1
(4) -8A = -1
(4) gives A = 1/8
(1) gives B = -1/8
(2) gives C = 1/4
The correct answer is
A = -1/4
B = 1/4
C = 2/4
What's wrong?
How can I expand this fraction?
\(\frac{A}{s} + \frac{B}{(s-2)} + \frac{C}{(s-2)^2}\)
right?
This gives me the equation
\(As^3 - 6As^2 + 12As - 8A Bs^3 - 4Bs^2 + 4Bs + Cs^2 - 2Cs = s-1\)
so that
(1) A + B =0
(2)- 6A - 4B + C = 0
(3) 12A + 4B - 2C = 1
(4) -8A = -1
(4) gives A = 1/8
(1) gives B = -1/8
(2) gives C = 1/4
The correct answer is
A = -1/4
B = 1/4
C = 2/4
What's wrong?
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