Final answer:
The polynomial 2x^3-8x^2+4x has three zeros. Factoring out the common factor 2x, we get a quadratic equation, which yields two additional zeros using the quadratic formula. Each zero (0, 2 + √2, 2 - √2) has a multiplicity of 1.
Step-by-step explanation:
To find the multiplicity of each zero in the polynomial function 2x^3-8x^2+4x, we can factorize the polynomial. Let's start by taking out the common factor, which is 2x:
2x(x^2 - 4x + 2).
The quadratic part doesn't factor over the integers, but we can use the quadratic formula to find its zeros:
x = [-(-4) ± √((-4)^2-4*1*2)]/(2*1)
x = [4 ± √(16-8)]/2
x = [4 ± √8]/2
x = [4 ± 2√2]/2
x = 2 ± √2
Therefore, the two zeros found using the quadratic formula are 2 + √2 and 2 - √2, both with multiplicity 1. The zero from the common factor 2x is x = 0 with multiplicity 1. Thus, we have three distinct zeros each with multiplicity 1.