Decide what values of the variable cannot possibly be solutions for the equation. Do not solve. \( \frac{1}{x-4}+\frac{1}{x+2}=\frac{1}{x^{2}-2 x-8} \) What values of \( x \) cannot be solutions of the equation? \( \square \) (Use a comma to separate answers as needed.)

To determine the values of \( x \) that cannot be solutions for the equation \[ \frac{1}{x-4}+\frac{1}{x+2}=\frac{1}{x^{2}-2 x-8}, \] we need to identify the values that would make any denominator in the equation equal to zero, as division by zero is undefined. 1. **Identify the denominators:** - The left side has two denominators: \( x - 4 \) and \( x + 2 \). - The right side has the denominator \( x^2 - 2x - 8 \). 2. **Set each denominator to zero and solve for \( x \):** - For \( x - 4 = 0 \): \[ x = 4 \] - For \( x + 2 = 0 \): \[ x = -2 \] - For \( x^2 - 2x - 8 = 0 \), we can factor it: \[ x^2 - 2x - 8 = (x - 4)(x + 2) = 0 \] This gives us the same solutions: \[ x = 4 \quad \text{and} \quad x = -2 \] 3. **Conclusion:** The values of \( x \) that cannot be solutions of the equation are \( 4 \) and \( -2 \). Thus, the final answer is: \( 4, -2 \)