all real zeros of the function. $$ f(x)=-2 x\left(x^{2}-4\right)\left(x^{2}+16\right) $$ re is more than one answer, separate them with commas.

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The function is 
$$
f(x)=-2 x(x^2-4)(x^2+16).
$$
To find the real zeros, we set $$ f(x)=0 $$.

1. Set each factor equal to zero:

- For $$ -2 $$, it never equals zero.
   
   - For $$ x $$:
     $$
     x = 0.
     $$

- For $$ x^2-4 $$:
     $$
     x^2-4 = 0 \quad \Longrightarrow \quad x^2 = 4 \quad \Longrightarrow \quad x = 2 \quad \text{or} \quad x = -2.
     $$

- For $$ x^2+16 $$:
     $$
     x^2+16 = 0 \quad \Longrightarrow \quad x^2 = -16.
     $$
     There are no real solutions since the square of a real number cannot be negative.

2. The real zeros are:
$$
-2, \ 0, \ 2.
$$
The real zeros are -2, 0, and 2.

all real zeros of the function.

re is more than one answer, separate them with commas.

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all real zeros of the function. re is more than one answer, separate them with commas.