How do you factor completely 16x^2-81?

Because there is no $x$ term and the constant term is a negative we know the sum of the factored $x$ terms is $0$. And, because both the ${x}^{2}$ coefficient and the constant are squares this expression can be factored as:
$\left(a x + b\right) \left(a x - b\right)$ or
$\left(4 x + 9\right) \left(4 x - 9\right)$