Points: 0.25 of 1 Save Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses the $x$-axis or touches the $x$-axis and turns around at each zero. \[ f(x)=4(x-2)(x+8)^{3} \] Determine the zero(s). The zero(s) is/are $-8,2$ (Type integers or decimals. Use a comma to separate answers as needed.) Determine the multiplicities of the zero(s). Select the correct choice below and, if necessary, fill in the answer box(es) within your choice. A. There is one zero. The multiplicity of the zero is $\square$. (Simplify your answer.) B. There are two zeros. The multiplicity of the largest zero is $\square$. The multiplicity of the smallest zero is $\square$ (Simplify your answers) C. There are three zeros. The multiplicity of the largest zero is $\square$. The multiplicity of the smallest zero is $\square$ The multiplicity of the other zero is $\square$ (Simplify your answers)

Step 1 ：The zeros of a polynomial function are the values of x that make the function equal to zero. In this case, the function is given in factored form, so the zeros are the values of x that make each factor equal to zero.

Step 2 ：The function is \(f(x)=4(x-2)(x+8)^{3}\). The zeros are the values of x that make each factor equal to zero, so \(x-2=0\) and \(x+8=0\). Solving these equations gives \(x=2\) and \(x=-8\).

Step 3 ：The multiplicity of a zero is the number of times that factor appears in the function. If a factor is raised to a power, that power is the multiplicity of the zero. The multiplicity of the zero \(x=2\) is 1, because the factor \((x-2)\) appears once. The multiplicity of the zero \(x=-8\) is 3, because the factor \((x+8)\) is raised to the power of 3.

Step 4 ：The graph of the function crosses the x-axis at a zero if the multiplicity of that zero is odd, and touches the x-axis and turns around at a zero if the multiplicity of that zero is even. The graph of the function crosses the x-axis at \(x=2\) because the multiplicity of that zero is odd, and touches the x-axis and turns around at \(x=-8\) because the multiplicity of that zero is even.

Step 5 ：\(\boxed{\text{Final Answer: The zeros are } x=2 \text{ and } x=-8. \text{ The multiplicity of the zero } x=2 \text{ is 1 and the multiplicity of the zero } x=-8 \text{ is 3. The graph of the function crosses the x-axis at } x=2 \text{ and touches the x-axis and turns around at } x=-8.}\)