Suppose
\(\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\)
where A, B, and C are real constants. What is A?
Multiplying the entire equation by \(x^3 - x^2 - 21x + 45\): \(1 = A(x-3)(x-3) + B(x+5)(x-3) + C(x-5)\)
Simplify the right-hand side: \(1 = A(x^2 - 6x + 9) + B(x^2 + 2x - 15) + C(x-5)\)
Equating terms to get the following system: \(Ax^2 + Bx^2 = 0\)
\(-6Ax + 2Bx + Cx = 0\)
\(9A - 15B - 5C = 1\)
Removing the x's from the system gives us a simpler system: \(A + B = 0\)
\(-6A + 2B + C = 0\)
\(9A - 15B - 5C = 1\)
Now, just solve for a.
:)