1. **State the problem:** Resolve the expression $$\frac{3x^2 - x + 8}{(x+4)(x^2+4)}$$ into partial fractions. 2. **Set up the partial fractions:** Since the denominator has a linear factor $(x+4)$ and an irreducible quadratic factor $(x^2+4)$, the decomposition is: $$\frac{3x^2 - x + 8}{(x+4)(x^2+4)} = \frac{A}{x+4} + \frac{Bx + C}{x^2 + 4}$$ where $A$, $B$, and $C$ are constants to be determined. 3. **Multiply both sides by the denominator** to clear fractions: $$3x^2 - x + 8 = A(x^2 + 4) + (Bx + C)(x + 4)$$ 4. **Expand the right side:** $$A x^2 + 4A + Bx^2 + 4Bx + Cx + 4C$$ Group like terms: $$ (A + B) x^2 + (4B + C) x + (4A + 4C) $$ 5. **Equate coefficients of corresponding powers of $x$:** - Coefficient of $x^2$: $3 = A + B$ - Coefficient of $x$: $-1 = 4B + C$ - Constant term: $8 = 4A + 4C$ 6. **Solve the system:** From the first equation: $$B = 3 - A$$ Substitute into the second: $$-1 = 4(3 - A) + C = 12 - 4A + C \implies C = -1 - 12 + 4A = 4A - 13$$ Substitute $C$ into the third: $$8 = 4A + 4(4A - 13) = 4A + 16A - 52 = 20A - 52$$ Add 52 to both sides: $$8 + 52 = 20A \implies 60 = 20A \implies A = 3$$ 7. **Find $B$ and $C$:** $$B = 3 - A = 3 - 3 = 0$$ $$C = 4A - 13 = 4(3) - 13 = 12 - 13 = -1$$ 8. **Write the final partial fraction decomposition:** $$\frac{3x^2 - x + 8}{(x+4)(x^2+4)} = \frac{3}{x+4} + \frac{0 \cdot x - 1}{x^2 + 4} = \frac{3}{x+4} - \frac{1}{x^2 + 4}$$