Find all real zeros of the function. \[ f(x)=2\left(x^{2}+25\right)(x-4)(x+2)^{2} \] If there is more than one answer, separate them with commas.

To find the real zeros of the function \( f(x) = 2\left(x^{2}+25\right)(x-4)(x+2)^{2} \), we need to set the function equal to zero: \[ f(x) = 0 \] The product of the factors equal to zero indicates that any factor can be solved for zero: 1. \( 2 \) is a constant and it doesn't affect the zeros. 2. The first factor \( x^2 + 25 = 0 \): \[ x^2 = -25 \] This gives us complex solutions \( x = 5i \) and \( x = -5i \), but there are no real zeros from this factor. 3. The second factor \( x - 4 = 0 \): \[ x = 4 \] 4. The third factor \( (x + 2)^2 = 0 \): \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] Note that \( (x + 2)^2 \) gives a double root, so it is counted as one zero but it appears twice. Thus, the real zeros of the function are: \[ \boxed{4, -2} \]