I think you mean 2m2(10m4-3m2+4) because the AC method can't be used on cubics. Although this revised expression contains a quartic (degree 4), it can be treated as a quadratic because only the even powers of m are present. In this case a=10 and c=4, while b=-3.
The factors of 10 are: (1,10), (-1,-10), (2,5), (-2, -5) and the factors of 4 are: (1,4), (-1,-4), (2,2), (-2,-2).
ac=40 and the factors of 40 are (1,40), (-1,-40), (2,20), (-2,-20), (4,10), (-4,-10), (5,8), (-5,-8).
We are looking for the pair of factors whose sum is -3. Clearly there are none, but, hey, the difference between 8 and 5 is 3, which is the size of b, so this suggests that the question has another error:
2m2(10m4-3m2+4) should be 2m2(10m4-3m2-4), making ac=-40 and corresponding adjustments to the factors, which now include (-5,8) and (5,-8). 5+(-8)=-3 which is b.
So we can now factor:
2m2(10m4+5m2-8m2-4)=2m2(5m2(2m2+1)-4(2m2+1))=2m2(5m2-4)(2m2+1). Further factorisation is possible but would generate irrational and complex factors.