1. **State the problem:** Decompose the rational function $$\frac{4x^2 - 10x - 6}{(x + 2)(x - 1)(3x + 1)}$$ into partial fractions. 2. **Formula and rules:** For distinct linear factors in the denominator, the partial fraction decomposition has the form: $$\frac{P(x)}{(x+a)(x+b)(x+c)} = \frac{A}{x+a} + \frac{B}{x+b} + \frac{C}{x+c}$$ where $A$, $B$, and $C$ are constants to be found. 3. **Set up the equation:** $$\frac{4x^2 - 10x - 6}{(x + 2)(x - 1)(3x + 1)} = \frac{A}{x + 2} + \frac{B}{x - 1} + \frac{C}{3x + 1}$$ Multiply both sides by the denominator to clear fractions: $$4x^2 - 10x - 6 = A(x - 1)(3x + 1) + B(x + 2)(3x + 1) + C(x + 2)(x - 1)$$ 4. **Expand each term:** - $A(x - 1)(3x + 1) = A(3x^2 + x - 3x - 1) = A(3x^2 - 2x - 1)$ - $B(x + 2)(3x + 1) = B(3x^2 + x + 6x + 2) = B(3x^2 + 7x + 2)$ - $C(x + 2)(x - 1) = C(x^2 - x + 2x - 2) = C(x^2 + x - 2)$ 5. **Combine:** $$4x^2 - 10x - 6 = A(3x^2 - 2x - 1) + B(3x^2 + 7x + 2) + C(x^2 + x - 2)$$ 6. **Group like terms:** $$4x^2 - 10x - 6 = (3A + 3B + C)x^2 + (-2A + 7B + C)x + (-A + 2B - 2C)$$ 7. **Equate coefficients:** - For $x^2$: $4 = 3A + 3B + C$ - For $x$: $-10 = -2A + 7B + C$ - Constant: $-6 = -A + 2B - 2C$ 8. **Solve the system:** From the first equation: $$C = 4 - 3A - 3B$$ Substitute into the second: $$-10 = -2A + 7B + (4 - 3A - 3B) = -5A + 4B + 4$$ Simplify: $$-14 = -5A + 4B$$ Substitute $C$ into the third: $$-6 = -A + 2B - 2(4 - 3A - 3B) = -A + 2B - 8 + 6A + 6B = 5A + 8B - 8$$ Simplify: $$2 = 5A + 8B$$ Now solve the system: $$\begin{cases} -14 = -5A + 4B \\ 2 = 5A + 8B \end{cases}$$ Add the two equations: $$-14 + 2 = (-5A + 5A) + (4B + 8B) \Rightarrow -12 = 12B \Rightarrow B = -1$$ Substitute $B = -1$ into $2 = 5A + 8B$: $$2 = 5A + 8(-1) = 5A - 8 \Rightarrow 5A = 10 \Rightarrow A = 2$$ Find $C$: $$C = 4 - 3(2) - 3(-1) = 4 - 6 + 3 = 1$$ 9. **Final answer:** $$\frac{4x^2 - 10x - 6}{(x + 2)(x - 1)(3x + 1)} = \frac{2}{x + 2} - \frac{1}{x - 1} + \frac{1}{3x + 1}$$