Is this correctly done?

Thanks Alan, I just explain a little further,

$$\\(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2\\\ so \boxed{a^2-b^2=(a-b)(a+b)}\\\\ \mbox{This is referred to as a difference of two squares and you need to memorize it.}\\\\ \begin{array}{rll} 16x^2-1&=&(4x)^2-1^2\\\\ &=&(4x-1)(4x+1) \end{array}$$

Yes, though you can take it a little further to get:  4x^2(4x+1)(4x-1)

So the factors are: 4, x^2, 4x+1 and 4x-1 (and various combinations of these).