1. **State the problem:** Express the rational function $$\frac{2x^3 - 4}{(x+3)(x-1)}$$ in partial fractions. 2. **Understand the problem:** Partial fraction decomposition requires the numerator degree to be less than the denominator degree. Here, numerator degree is 3 and denominator degree is 2 (since $(x+3)(x-1)$ expands to $x^2 + 2x - 3$). 3. **Divide numerator by denominator:** Perform polynomial division to rewrite the expression as a polynomial plus a proper fraction. Divide $2x^3 - 4$ by $x^2 + 2x - 3$: - Leading term division: $\frac{2x^3}{x^2} = 2x$ - Multiply divisor by $2x$: $2x(x^2 + 2x - 3) = 2x^3 + 4x^2 - 6x$ - Subtract: $(2x^3 - 4) - (2x^3 + 4x^2 - 6x) = -4x^2 + 6x - 4$ Next term: - Leading term division: $\frac{-4x^2}{x^2} = -4$ - Multiply divisor by $-4$: $-4(x^2 + 2x - 3) = -4x^2 - 8x + 12$ - Subtract: $(-4x^2 + 6x - 4) - (-4x^2 - 8x + 12) = 14x - 16$ So, $$\frac{2x^3 - 4}{(x+3)(x-1)} = 2x - 4 + \frac{14x - 16}{(x+3)(x-1)}$$ 4. **Set up partial fractions for the remainder:** $$\frac{14x - 16}{(x+3)(x-1)} = \frac{A}{x+3} + \frac{B}{x-1}$$ Multiply both sides by $(x+3)(x-1)$: $$14x - 16 = A(x - 1) + B(x + 3)$$ 5. **Find A and B by substituting convenient values:** - For $x = 1$: $$14(1) - 16 = A(1 - 1) + B(1 + 3) \Rightarrow -2 = 4B \Rightarrow B = -\frac{1}{2}$$ - For $x = -3$: $$14(-3) - 16 = A(-3 - 1) + B(-3 + 3) \Rightarrow -42 - 16 = -4A + 0 \Rightarrow -58 = -4A \Rightarrow A = \frac{29}{2}$$ 6. **Write the final partial fraction decomposition:** $$\frac{2x^3 - 4}{(x+3)(x-1)} = 2x - 4 + \frac{\frac{29}{2}}{x+3} - \frac{\frac{1}{2}}{x-1}$$ Or equivalently: $$\frac{2x^3 - 4}{(x+3)(x-1)} = 2x - 4 + \frac{29}{2(x+3)} - \frac{1}{2(x-1)}$$