Question 6256
You did the first step right--you factored out the common factor of 2:
{{{2m^2n^2 - 32mn^2 + 8m^2 - 128}}}
{{{2(m^2n^2 - 16mn^2 + 4m^2 - 64)}}}


However what remains is something with 4 terms that factors by "grouping."  From the first two terms, take out an {{{n^2}}}, and from the last two terms, take out a 4.
{{{2(n^2(m^2 - 16m) + 4(m^2 - 16) )}}}


At this point, I suspect that you copied the problem right.  In order to factor by grouping, you MUST have a common factor after you do the grouping and factor as we did here.  As you can see, this is NOT the case, because of the extra m in the first grouping.  Let me suggest what I think the problem should have been:

{{{2m^3n^2 - 32mn^2 + 8m^2 - 128}}}
{{{2(m^3n^2 - 16mn^2 + 4m^2 - 64)}}}
{{{2(mn^2(m^2 - 16) + 4(m^2 - 16) )}}}


Now you can take out the common factor, which is {{{(m^2 - 16)}}}
{{{2(m^2-16) (mn^2 +4) }}}


Last, factor the {{{m^2 - 16}}} which is the difference of two squares:
{{{2(m-4)(m+4) (mn^2+4)}}}


R^2 at SCC