1. **State the problem:** Decompose the rational expression $$\frac{3x^2 + 4}{(x-3)(x^2 + 2)}$$ into partial fractions. 2. **Formula and rules:** For a rational expression where the denominator factors into linear and irreducible quadratic terms, the partial fraction decomposition has the form: $$\frac{P(x)}{(x-3)(x^2+2)} = \frac{A}{x-3} + \frac{Bx + C}{x^2 + 2}$$ where $A$, $B$, and $C$ are constants to be determined. 3. **Set up the equation:** Multiply both sides by the denominator $(x-3)(x^2+2)$ to clear fractions: $$3x^2 + 4 = A(x^2 + 2) + (Bx + C)(x - 3)$$ 4. **Expand the right side:** $$A x^2 + 2A + Bx^2 - 3Bx + Cx - 3C$$ Group like terms: $$ (A + B) x^2 + (-3B + C) x + (2A - 3C) $$ 5. **Equate coefficients:** From the left side $3x^2 + 0x + 4$, match coefficients: - For $x^2$: $A + B = 3$ - For $x$: $-3B + C = 0$ - For constant: $2A - 3C = 4$ 6. **Solve the system:** From $-3B + C = 0$, we get $C = 3B$. Substitute into $2A - 3C = 4$: $$2A - 3(3B) = 4 \Rightarrow 2A - 9B = 4$$ From $A + B = 3$, we have $A = 3 - B$. Substitute $A$ into the above: $$2(3 - B) - 9B = 4 \Rightarrow 6 - 2B - 9B = 4 \Rightarrow 6 - 11B = 4$$ $$-11B = -2 \Rightarrow B = \frac{2}{11}$$ Then $A = 3 - \frac{2}{11} = \frac{33}{11} - \frac{2}{11} = \frac{31}{11}$. And $C = 3B = 3 \times \frac{2}{11} = \frac{6}{11}$. 7. **Write the decomposition:** $$\frac{3x^2 + 4}{(x-3)(x^2 + 2)} = \frac{31/11}{x-3} + \frac{(2/11)x + 6/11}{x^2 + 2}$$ **Final answer:** $$\frac{3x^2 + 4}{(x-3)(x^2 + 2)} = \frac{31}{11(x-3)} + \frac{2x + 6}{11(x^2 + 2)}$$