1. **State the problem:** Express the rational function $\frac{2x+3}{(x-1)(2x+1)(x-3)}$ as a sum of partial fractions. 2. **Formula and rules:** For distinct linear factors in the denominator, the partial fraction decomposition is: $$\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}$$ where $A$, $B$, and $C$ are constants to be found. 3. **Set up the decomposition:** $$\frac{2x+3}{(x-1)(2x+1)(x-3)} = \frac{A}{x-1} + \frac{B}{2x+1} + \frac{C}{x-3}$$ 4. **Multiply both sides by the denominator to clear fractions:** $$2x+3 = A(2x+1)(x-3) + B(x-1)(x-3) + C(x-1)(2x+1)$$ 5. **Expand each term:** - $A(2x+1)(x-3) = A(2x^2 -6x + x -3) = A(2x^2 -5x -3)$ - $B(x-1)(x-3) = B(x^2 -4x +3)$ - $C(x-1)(2x+1) = C(2x^2 + x -2x -1) = C(2x^2 - x -1)$ 6. **Rewrite the equation:** $$2x + 3 = A(2x^2 -5x -3) + B(x^2 -4x +3) + C(2x^2 - x -1)$$ 7. **Group like terms:** $$2x + 3 = (2A + B + 2C)x^2 + (-5A -4B - C)x + (-3A + 3B - C)$$ 8. **Equate coefficients of powers of $x$ on both sides:** - Coefficient of $x^2$: $0 = 2A + B + 2C$ - Coefficient of $x$: $2 = -5A -4B - C$ - Constant term: $3 = -3A + 3B - C$ 9. **Solve the system:** From $0 = 2A + B + 2C$, express $B = -2A - 2C$. Substitute into the other equations: - $2 = -5A -4(-2A - 2C) - C = -5A + 8A + 8C - C = 3A + 7C$ - $3 = -3A + 3(-2A - 2C) - C = -3A -6A -6C - C = -9A -7C$ 10. **Rewrite system:** $$\begin{cases} 2 = 3A + 7C \\ 3 = -9A -7C \end{cases}$$ Add the two equations: $$2 + 3 = 3A + 7C - 9A - 7C = -6A \implies 5 = -6A \implies A = -\frac{5}{6}$$ 11. **Find $C$:** From $2 = 3A + 7C$: $$2 = 3\left(-\frac{5}{6}\right) + 7C = -\frac{15}{6} + 7C = -\frac{5}{2} + 7C$$ $$7C = 2 + \frac{5}{2} = \frac{4}{2} + \frac{5}{2} = \frac{9}{2} \implies C = \frac{9}{14}$$ 12. **Find $B$:** $$B = -2A - 2C = -2\left(-\frac{5}{6}\right) - 2\left(\frac{9}{14}\right) = \frac{10}{6} - \frac{18}{14} = \frac{5}{3} - \frac{9}{7} = \frac{35}{21} - \frac{27}{21} = \frac{8}{21}$$ 13. **Final answer:** $$\frac{2x+3}{(x-1)(2x+1)(x-3)} = \frac{-\frac{5}{6}}{x-1} + \frac{\frac{8}{21}}{2x+1} + \frac{\frac{9}{14}}{x-3} = -\frac{5}{6(x-1)} + \frac{8}{21(2x+1)} + \frac{9}{14(x-3)}$$