1. **Express the partial fraction decomposition of** $\frac{3x}{(x+1)(x-2)}$. 2. **State the problem:** Decompose $\frac{3x}{(x+1)(x-2)}$ into partial fractions. 3. **Formula:** For distinct linear factors, write $$\frac{3x}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$$ where $A$ and $B$ are constants to find. 4. **Multiply both sides by the denominator:** $$3x = A(x-2) + B(x+1)$$ 5. **Expand the right side:** $$3x = Ax - 2A + Bx + B$$ 6. **Group like terms:** $$3x = (A + B)x + (-2A + B)$$ 7. **Equate coefficients:** - Coefficient of $x$: $3 = A + B$ - Constant term: $0 = -2A + B$ 8. **Solve the system:** From $0 = -2A + B$, we get $B = 2A$. Substitute into $3 = A + B$: $$3 = A + 2A = 3A \implies A = 1$$ Then $B = 2(1) = 2$. 9. **Final partial fraction decomposition:** $$\frac{3x}{(x+1)(x-2)} = \frac{1}{x+1} + \frac{2}{x-2}$$ **Answer:** $\boxed{\frac{3x}{(x+1)(x-2)} = \frac{1}{x+1} + \frac{2}{x-2}}$