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 $x = \frac{3}{4}$ .

Answer

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Answer

\boxed{(A,B,C) = (2,2,-3)}\)

Steps

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

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

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 $4 x^{2} - 4 x - 9 = A (2 x + 1) (x - 1) - 3 (2 x + 1) + 2 (x - 1)$

Step 6 ：Comparing coefficients, we have $A = 2$

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

Step 8 ：\boxed{(A,B,C) = (2,2,-3)}\)