1. **Problem:** Find the values of A, B, and C such that $$\frac{2x+1}{(x+1)(x+2)^2} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$ 2. **Formula and rules:** For partial fraction decomposition, express the rational function as a sum of simpler fractions with unknown coefficients. Multiply both sides by the denominator to clear fractions: $$2x+1 = A(x+2)^2 + B(x+1)(x+2) + C(x+1)$$ 3. **Expand terms:** $$A(x^2 + 4x + 4) + B(x^2 + 3x + 2) + C(x+1)$$ 4. **Combine like terms:** $$Ax^2 + 4Ax + 4A + Bx^2 + 3Bx + 2B + Cx + C$$ Group by powers of $x$: $$ (A + B)x^2 + (4A + 3B + C)x + (4A + 2B + C)$$ 5. **Equate coefficients with left side $2x + 1$:** - Coefficient of $x^2$: $A + B = 0$ - Coefficient of $x$: $4A + 3B + C = 2$ - Constant term: $4A + 2B + C = 1$ 6. **Solve system:** From $A + B = 0$, we get $B = -A$. Substitute into other equations: - $4A + 3(-A) + C = 2 \Rightarrow 4A - 3A + C = 2 \Rightarrow A + C = 2$ - $4A + 2(-A) + C = 1 \Rightarrow 4A - 2A + C = 1 \Rightarrow 2A + C = 1$ Subtract second from first: $A + C - (2A + C) = 2 - 1 \Rightarrow -A = 1 \Rightarrow A = -1$ Then $B = -A = 1$. From $A + C = 2$, $-1 + C = 2 \Rightarrow C = 3$. 7. **Answer:** $A = -1$, $B = 1$, $C = 3$ which corresponds to option C. **Final answer:** C) -1, 1, 3