2. $\frac{4 x^{2}-4 x-9}{(2 x+1)(x-1)} \equiv A+\frac{B}{2 x+1}+\frac{C}{x-1}$ a Find the values of the constants $A, B$ and $C$. b Hence, or otherwise, expand $\frac{4 x^{2}-4 x-9}{(2 x+1)(x-1)}$ in ascending powers of $x$, as far as the $x^{2}$ term c Explain why the expansion is not valid for $x=\frac{3}{4}$.

Step 1 ：\(\frac{4x^2-4x-9}{(2x+1)(x-1)} = A + \frac{B}{2x+1} + \frac{C}{x-1}\)

Step 2 ：Multiplying by \((2x+1)(x-1)\) gives \(4x^2-4x-9 = A(2x+1)(x-1) + B(x-1) + C(2x+1)\)

Step 3 ：Let \(x = 1\), then \(4(1)^2 - 4(1) - 9 = C(2(1) + 1)\), so \(C = -3\)

Step 4 ：Let \(x = -\frac{1}{2}\), then \(4(-\frac{1}{2})^2 - 4(-\frac{1}{2}) - 9 = B(-\frac{1}{2} - 1)\), so \(B = 2\)

Step 5 ：Substitute \(B = 2\) and \(C = -3\) into the equation, we get \(4x^2 - 4x - 9 = A(2x+1)(x-1) - 3(2x+1) + 2(x-1)\)

Step 6 ：Comparing coefficients, we have \(A = 2\)

Step 7 ：\(\frac{4x^2-4x-9}{(2x+1)(x-1)} = 2 + \frac{2}{2x+1} - \frac{3}{x-1}\)

Step 8 ：\boxed{(A,B,C) = (2,2,-3)}\)