1. **State the problem:** We want to express the function $$f(x) = \frac{9x^2 + 4}{(2x + 1)(x - 2)^2}$$ as a sum of partial fractions. 2. **Recall the formula and rules:** For a rational function where the denominator factors into linear and repeated linear factors, the partial fraction decomposition takes the form: $$\frac{P(x)}{(ax + b)(x - c)^n} = \frac{A}{ax + b} + \frac{B}{x - c} + \frac{C}{(x - c)^2} + \cdots$$ Here, since $(x-2)$ is squared, we include terms for both powers. 3. **Set up the partial fractions:** $$\frac{9x^2 + 4}{(2x + 1)(x - 2)^2} = \frac{A}{2x + 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}$$ 4. **Multiply both sides by the denominator to clear fractions:** $$9x^2 + 4 = A(x - 2)^2 + B(2x + 1)(x - 2) + C(2x + 1)$$ 5. **Expand each term:** - $(x - 2)^2 = x^2 - 4x + 4$ - $(2x + 1)(x - 2) = 2x^2 - 3x - 2$ So, $$9x^2 + 4 = A(x^2 - 4x + 4) + B(2x^2 - 3x - 2) + C(2x + 1)$$ 6. **Distribute constants:** $$9x^2 + 4 = A x^2 - 4 A x + 4 A + 2 B x^2 - 3 B x - 2 B + 2 C x + C$$ 7. **Group like terms:** $$9x^2 + 4 = (A + 2 B) x^2 + (-4 A - 3 B + 2 C) x + (4 A - 2 B + C)$$ 8. **Equate coefficients of powers of $x$:** - Coefficient of $x^2$: $9 = A + 2 B$ - Coefficient of $x$: $0 = -4 A - 3 B + 2 C$ - Constant term: $4 = 4 A - 2 B + C$ 9. **Solve the system:** From the first equation: $$A = 9 - 2 B$$ Substitute into the second: $$0 = -4(9 - 2 B) - 3 B + 2 C = -36 + 8 B - 3 B + 2 C = -36 + 5 B + 2 C$$ Rearranged: $$2 C = 36 - 5 B \implies C = \frac{36 - 5 B}{2}$$ Substitute $A$ and $C$ into the third: $$4 = 4(9 - 2 B) - 2 B + \frac{36 - 5 B}{2} = 36 - 8 B - 2 B + \frac{36 - 5 B}{2}$$ Multiply both sides by 2 to clear denominator: $$8 = 72 - 20 B + 36 - 5 B = 108 - 25 B$$ Rearranged: $$25 B = 108 - 8 = 100 \implies B = 4$$ Then: $$A = 9 - 2(4) = 9 - 8 = 1$$ $$C = \frac{36 - 5(4)}{2} = \frac{36 - 20}{2} = \frac{16}{2} = 8$$ 10. **Write the final partial fraction decomposition:** $$f(x) = \frac{1}{2x + 1} + \frac{4}{x - 2} + \frac{8}{(x - 2)^2}$$