2m^2(10m^5-3m^2+4)

Question

factor the problem using the AC method

Answer 1

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.