You stated it incorrectly, but essentially you're on the right track.

This is not an equation: (w - 8)(w + 12). You could mean either the solutions to the equation (w - 8)(w + 12) = 0 (note: that has an equals sign and expressions on both sides of the equals sign. That makes it an equation, a thing which potentially has a solution).

Or you could be asking for the roots of the expression (w - 8)(w + 12), which are the values where it is 0. So it's the same as the solution to the equation.

"Since (w - 8)" is not a valid statement either. What about w - 8? Are you saying the expression w - 8 is true? What would it mean for that to be true? If I say "x + 1" to you, can you tell me whether that's a thing which is true or false?

No, of course not. What you have heard is that if a product is ZERO, if (w - 8)(w + 12) = 0, then one of the factors must be 0. So EITHER w - 8 = 0 (note again: there's an equals sign, and something on both sides of the equals sign. That is a statement which can be true or false), or w + 12 = 0.

That's called the Zero Factor Theorem but if you think about it, it's kind of obvious. Anything times 0 is 0, and you can't multiply two nonzero things to get zero. So if two things multiply to be 0, one or both of them must be 0.

So you remembered the method correctly but stated the reasoning incorrectly.

Edit: "That is, you take the number in each parenthesis and just put the opposite sign on them and that is your solutions for the equation?"

In effect, IN THIS CASE. But what you really do if the whole product is zero, is set each factor equal to 0 and then solve those equation.

Here are some examples where the way you stated the rule won't work:

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

Solutions: Either x = 0, or x - 3 = 0, or x^2 + 2 = 0. There is no real x that will make x^2 + 2 = 0.

(x^3 + 27)(x^2 - 4) = 0

Solutions: Either x^3 + 27 = 0 (there is one x that solves that), or x^2 - 4 = 0 (there are two x's that solve that)

Edit 2: As someone else pointed out, you might have an equation where the right-hand side is not 0. In that case, this won't work. If I told you (x + 1)(x - 2) = 5, that wouldn't mean x + 1 has to equal 5 or anything else in particular. I can't conclude anything about the factors from knowing they are two real numbers which multiply to make 5, except that both are nonzero.