Let's solve the partial fraction decomposition step by step for: \[ \frac{6x^2 - 14x - 27}{(x+2)(x-3)^2} \] --- ## **Step 1: Set Up the Partial Fractions** Since the denominator is \((x+2)(x-3)^2\), we write: \[ \frac{6x^2 - 14x - 27}{(x+2)(x-3)^2} = \frac{A}{x+2} + \frac{B}{x-3} + \frac{C}{(x-3)^2} \] --- ## **Step 2: Combine Right Side Over a Common Denominator** \[ \frac{A}{x+2} + \frac{B}{x-3} + \frac{C}{(x-3)^2} = \frac{A(x-3)^2 + B(x+2)(x-3) + C(x+2)}{(x+2)(x-3)^2} \] --- ## **Step 3: Set Numerators Equal** \[ 6x^2 - 14x - 27 = A(x-3)^2 + B(x+2)(x-3) + C(x+2) \] --- ## **Step 4: Expand the Right Side** Expand each term: - \(A(x-3)^2 = A(x^2 - 6x + 9)\) - \(B(x+2)(x-3) = B(x^2 - 3x + 2x - 6) = B(x^2 - x - 6)\) - \(C(x+2) = Cx + 2C\) So, \[ A(x^2 - 6x + 9) + B(x^2 - x - 6) + (Cx + 2C) \] \[ = (A + B)x^2 + (-6A - B + C)x + (9A - 6B + 2C) \] --- ## **Step 5: Match Coefficients** Set equal to the original numerator: \[ 6x^2 - 14x - 27 = (A + B)x^2 + (-6A - B + C)x + (9A - 6B + 2C) \] So, - \(x^2\) terms: \(A + B = 6\) - \(x\) terms: \(-6A - B + C = -14\) - Constant terms: \(9A - 6B + 2C = -27\) --- ## **Step 6: Solve the System of Equations** ### (1) \(A + B = 6\) ### (2) \(-6A - B + C = -14\) ### (3) \(9A - 6B + 2C = -27\) #### Substitute \(B = 6 - A\) into (2) and (3): From (2): \[ -6A - (6 - A) + C = -14 \\ -6A - 6 + A + C = -14 \\ -5A + C = -8 \\ C = 5A - 8 \tag{4} \] From (3): \[ 9A - 6(6 - A) + 2C = -27 \\ 9A - 36 + 6A + 2C = -27 \\ 15A + 2C = 9 \\ 2C = 9 - 15A \\ C = \frac{9 - 15A}{2} \tag{5} \] Set equation (4) and (5) for \(C\) equal: \[ 5A - 8 = \frac{9 - 15A}{2} \] Multiply both sides by 2: \[ 10A - 16 = 9 - 15A \\ 10A + 15A = 9 + 16 \\ 25A = 25 \\ A = 1 \] Now, \(B = 6 - A = 5\) Now, \(C = 5A - 8 = 5(1) - 8 = -3\) --- ## **Step 7: Write the Final Answer** \[ \frac{6x^2 - 14x - 27}{(x+2)(x-3)^2} = \frac{1}{x+2} + \frac{5}{x-3} - \frac{3}{(x-3)^2} \] --- ### **Final Answer** \[ \boxed{ \frac{6x^2 - 14x - 27}{(x+2)(x-3)^2} = \frac{1}{x+2} + \frac{5}{x-3} - \frac{3}{(x-3)^2} } \]