Erm... okay. I'm still a bit confused, but I think I understand what's going on. There are actually two completely separate concepts you're asking about, and they're really not related at all. The domain of a function is the set of all values the variable can take on. The zero product property has no role in determining this whatsoever. Using your example function:
\(\displaystyle f(x)=\dfrac{14x}{3x+5}\)
The domain is the set of all inputs (i.e. x-values) for the function. Because there's an infinite amount of numbers, it's often easier to determine which inputs are invalid, rather than determine which are valid. You first used the inequality \(\displaystyle 3x+5 \ge 0\) but I'm not sure what this is for. That inequality establishes what x values produce a negative denominator, but it doesn't matter if the denominator is negative. As an example, if x=1, the function is \(\displaystyle f(x)=\dfrac{14(1)}{3(1)-5}=\dfrac{14}{-2}=-7\). The denominator is negative, but the function is still defined. So what can you conclude about if x=1 is part of the domain? What value(s) of x make the function undefined? Are those x-values part of the domain?
Next, we have the zero product principle. This is used for determining roots (sometimes called solutions, or zeroes of a function, not the domain. In your example:
\(\displaystyle f(x)=x^2+7x+10\)
The roots of this function are when the function equals 0. So you would set it equal, and factor, as you did. Essentially, all the zero product principle says is this:
If "something" times "something else" equals 0, then we know that either "something" is 0, or "something else" is 0. Do you see why that must be the case?