Find all zeros (real and complex) of the polynomial. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities \[ P(x)=x^{4}-2 x^{3}-7 x^{2}-2 x-8 \] \( x=\square \)

To find the zeros of the polynomial \( P(x) = x^{4} - 2x^{3} - 7x^{2} - 2x - 8 \), we can apply the Rational Root Theorem first. Testing possible rational roots (factors of -8), we find \( x = 2 \) is a root. We then perform synthetic division to simplify \( P(x) \) into a cubic polynomial: \[ P(x) = (x - 2)(x^{3} + 2x^{2} - 3x - 4). \] Next, we can further factor \( x^{3} + 2x^{2} - 3x - 4 \). Testing \( x = -2 \) shows it is also a root. Again, perform synthetic division: \[ x^{3} + 2x^{2} - 3x - 4 = (x + 2)(x^{2} - 4). \] The quadratic \( x^{2} - 4 \) can be factored as \( (x - 2)(x + 2) \). Thus, we have: \[ P(x) = (x - 2)^2 (x + 2)^2. \] The complete list of zeros is: \[ x = 2 \text{ (multiplicity 2)}, -2 \text{ (multiplicity 2)}. \] So, the zeros in a comma-separated list are: \( 2, 2, -2, -2 \).