Express \( \frac{x^{2}+6 x-4}{(x+2)^{2}(x-4)} \) in partial fractions. Hence, or otherwise, obtain the first three non-zero terms in the Maclaurin expansion of \( \frac{x^{2}+6 x-4}{(x+2)^{2}(x-4)} \)

To express \( \frac{x^{2}+6 x-4}{(x+2)^{2}(x-4)} \) in partial fractions, we start by assuming the form: \[ \frac{x^{2}+6 x-4}{(x+2)^{2}(x-4)} = \frac{A}{x+2} + \frac{B}{(x+2)^{2}} + \frac{C}{x-4} \] Multiplying through by the denominator \( (x+2)^{2}(x-4) \), we have: \[ x^{2}+6x-4 = A(x+2)(x-4) + B(x-4) + C(x+2)^{2} \] Expanding the right-hand side: 1. \(A(x+2)(x-4) = A(x^2 - 4x + 2x - 8) = A(x^2 - 2x - 8)\) 2. \(B(x-4) = Bx - 4B\) 3. \((x+2)^{2} = x^{2} + 4x + 4\), thus \(C(x+2)^{2} = C(x^2 + 4x + 4)\) Combining: \[ Cx^2 + (A - 2A + B + 4C)x + (-8A - 4B + 4C) \] Setting coefficients equal to those on the left-hand side provides a system of equations: 1. \(A + C = 1\) (coefficient of \(x^2\)) 2. \(-2A + B + 4C = 6\) (coefficient of \(x\)) 3. \(-8A - 4B + 4C = -4\) (constant term) Solving these equations: From (1): \(C = 1 - A\). Substituting into (2): \[ -2A + B + 4(1 - A) = 6 \implies -6A + B + 4 = 6 \implies B = 6A + 2 \] Now, substituting \(B\) and \(C\) in equation (3): \[ -8A - 4(6A + 2) + 4(1 - A) = -4 \\ -8A - 24A - 8 + 4 - 4A = -4 \\ -36A - 4 = -4 \\ -36A = 0 \implies A = 0 \] From \(A = 0\): - \(C = 1\) - \(B = 2\) Thus, the partial fraction decomposition is: \[ \frac{0}{x+2} + \frac{2}{(x+2)^{2}} + \frac{1}{x-4} = \frac{2}{(x+2)^{2}} + \frac{1}{x-4} \] Next step, we obtain the first three non-zero terms in the Maclaurin expansion: The function breaks down as: 1. For \( \frac{2}{(x+2)^{2}} \): \[ \frac{2}{(2 + x)^{2}} = \frac{2}{4} \cdot (1 - \frac{x}{2})^{-2} \approx \frac{1}{2} \left( 1 + 2\frac{x}{2} + 3\left(\frac{x}{2}\right)^{2} + \cdots \right) = \frac{1}{2} \left( 1 + x + \frac{3x^2}{4} + \cdots \right) \] Results in \( \frac{1}{2} + \frac{x}{2} + \frac{3x^{2}}{8} + \cdots \) 2. For \( \frac{1}{x-4} \): \[ \frac{1}{-4(1 - \frac{x}{4})} \approx -\frac{1}{4}(1 + \frac{x}{4} + \cdots) = -\frac{1}{4} - \frac{x}{16} + \cdots \] Combining these expansions gives us: \[ \frac{1}{2} - \frac{1}{4} + \left( \frac{x}{2} - \frac{x}{16} \right) + \left( \frac{3x^{2}}{