Find all zeros of the function $f(x)=8(x-1)^{3}(3 x+5)\left(x^{2}-25\right)$ and their multiplicities.

Choose the zeros of $\mathrm{f}(\mathrm{x})$ and their multiplicities.
A. 1 (multiplicity 3 ), $-\frac{5}{3}, 5$ (multiplicity 2 )
B. $1, \frac{5}{3}, 5,-5$
C. 1 (multiplicity 3 ), $-\frac{5}{3}, 5,-5$
D. $-8,1$ (multiplicity 3$),-\frac{5}{3}, 5,-5$

Step 1 ：The zeros of the function are the values of x that make the function equal to zero. To find the zeros, we need to set the function equal to zero and solve for x. The multiplicity of a zero is the number of times it appears as a root of the function, which is determined by the exponent of the factor in the function.

Step 2 ：In this case, the function is already factored, so we can easily see the zeros and their multiplicities. The zeros are the values of x that make each factor equal to zero.

Step 3 ：The factor \((x-1)^{3}\) has a zero at x=1 with multiplicity 3, because the exponent of the factor is 3.

Step 4 ：The factor \((3x+5)\) has a zero at x=-5/3 with multiplicity 1, because the exponent of the factor is 1.

Step 5 ：The factor \((x^{2}-25)\) has zeros at x=5 and x=-5 with multiplicity 1, because the exponent of the factor is 1. However, since this factor is a quadratic, each zero actually has multiplicity 2, because the exponent of the x term in the quadratic is 2.

Step 6 ：So, the zeros of the function are x=1 with multiplicity 3, x=-5/3 with multiplicity 1, and x=5 and x=-5 each with multiplicity 2.

Step 7 ：Final Answer: \(\boxed{1 \text{ (multiplicity 3)}, -\frac{5}{3}, 5 \text{ (multiplicity 2)}, -5 \text{ (multiplicity 2)}}\)