Roots of denominator are -3+2j, -3-2j, does the final value theorem apply

Conditions of FVT are

1- All roots of the denominator of H(s) must have negative real parts.
2- H(s) must not have more than one pole at the origin.

http://en.wikipedia.org/wiki/Final_value…

Assuming you are in the same engineering process as whoever wrote the Wikipedia article, I think it applies. (As a mathematician I think there are other conditions unstated about H(s), such as being a complex analytical function say.)

Indeed -3+2j and -3-2j have negative real parts, which is -3 in both cases, and I believe -3 < 0.

Also if it had more than one pole at the origin then the denominator would have 0 as a root with multiplicity 2 or more. I do not see 0 in the list of roots of the denominator. Also apparently poles in the numerator are something that nobody expects to see (hopefully with good reason.)

Try to be somewhat careful, because we know what happened with Blackwater Horizon and the Verrazano Narrows etc.

I agree with what "vekkus4" has said.
The second condition should read that H(s) does not have a pole of order 2 or higher at z = 0.