b³+3b²-4b-12

Factor by grouping

Group the first two terms: b³+3b² factors as b²(b+3)

Factor the last two terms: -4b-12 factors as -4(b+3)

So b³+3b²-4b-12 equals b²(b+3)-4(b+3)

Take out the common factor (b+3) and get (b+3)(b²-4)

Factor the second parentheses as the difference of 
squares (b-2)(b+2), which means that

b³+3b²-4b-12 equals (b+3)(b-2)(b+2)

Therefore the dimensions are b+3, b-2, and b+2 where b > 2

Since these three factors are the dimensions of a box, they 
cannot be negative.  So b must be greater than 2 to prevent 
b-2 from being negative.

Edwin