find all real zeros and their multiplicity of the polynomial function

f(x)=5(x-4)squared(x+2)

To find the real zeros of a polynomial function, we need to find the values of x that make the function equal to zero.

In this case, we have the polynomial function f(x) = 5(x - 4)^2(x + 2).

To find the real zeros, we set f(x) equal to zero and solve for x:

5(x - 4)^2(x + 2) = 0

Since we have the product of three factors equal to zero, one or more of the factors must be zero.

Setting each factor equal to zero gives us three equations:

1) x - 4 = 0
2) (x - 4)^2 = 0
3) x + 2 = 0

Solving equation 1, we have:

x - 4 = 0
x = 4

So, x = 4 is a zero of multiplicity 1 (since it appears as a linear factor).

For equation 2, we have:

(x - 4)^2 = 0

Taking the square root of both sides, we get:

x - 4 = 0
x = 4

So, x = 4 is a zero of multiplicity 2 (since it appears as a squared factor).

Solving equation 3, we have:

x + 2 = 0
x = -2

So, x = -2 is a zero of multiplicity 1 (since it appears as a linear factor).

Therefore, the real zeros and their multiplicities for the polynomial function f(x) = 5(x - 4)^2(x + 2) are:
- Zero: x = 4
Multiplicity: 2
- Zero: x = -2
Multiplicity: 1