1. Express the partial fraction decomposition of $ \frac{3x}{(x+1)(x-2)} $. 2. The problem is to write $ \frac{3x}{(x+1)(x-2)} $ as a sum of simpler fractions of the form $ \frac{A}{x+1} + \frac{B}{x-2} $. 3. Set up the equation: $$\frac{3x}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$$ Multiply both sides by $ (x+1)(x-2) $ to clear denominators: $$3x = A(x-2) + B(x+1)$$ 4. Expand the right side: $$3x = A x - 2A + B x + B$$ Group like terms: $$3x = (A + B) x + (-2A + B)$$ 5. Equate coefficients of $ x $ and the constant terms: - Coefficient of $ x $: $3 = A + B$ - Constant term: $0 = -2A + B$ 6. Solve the system: From $0 = -2A + B$, we get $B = 2A$. Substitute into $3 = A + B$: $$3 = A + 2A = 3A \implies A = 1$$ Then $B = 2 \times 1 = 2$. 7. Therefore, the partial fraction decomposition is: $$\frac{3x}{(x+1)(x-2)} = \frac{1}{x+1} + \frac{2}{x-2}$$ Final answer: $$\boxed{\frac{3x}{(x+1)(x-2)} = \frac{1}{x+1} + \frac{2}{x-2}}$$