I'm confused about the process of how to find the degree of a polynomial factored function. I'm not sure that in this specific question if there is only two zeros/factors or if the 0.5x also plays into the degree of the function.
$(x+4)^2$ is of order of 2 and $(2x-3)$ Is order of 1 so I come to the conclusion this function is degree of 3.
What I'm not sure about is if because that leading coefficient has a x next to it does it make it a degree of 4?
I've also read that counting the amount of x's will give me the degree of a factored function is that right? And if so does the order of factor $(x+4)^2$ add to my degree?
The degree of the function is 4. You are right by saying $(x+4)^2$ is of degree 2 and $2x-3$ is of degree 1. $x$ is also of degree 1, and is not a coefficient.
In general, if $P(x) = f(x)\cdot g(x) \cdot h(x)$, then $\deg{P} = \deg{f} + \deg{g} + \deg{h}$.
Just counting the number of $x$'s you see in the function will not always give you the correct answer (not 3 in this case, it's 4). However, if you remember to count the $x$ in $(x+4)^2$ twice, then yes, counting the number of $x$'s will give you the degree.
If we were to expand the entire function out, one of the terms would be the $0.5x$ multiplied by the $x^2$ from the expansion of $(x+4)^2 = x^2 + 8x + 16$ multiplied by $2x$. Hence, we would have an $x^4$ term in our polynomial and hence the polynomial is said to be of degree 4.