Consecutive integers are 1 apart x,x+1,x+2 (x)(x+1)(x+2)=-120 x^3+3x^2+3x=-120 add 120 to both sides x^3+3x^2+3x+120=0 factor (x+6)(x^2-3x+20)=0 set each to zero
x+6=0 x=-6
x^2-3x+20=0 will yeild non-real result, discard x=-6 x+1=-5 x+2=-4 the numbers are -4,-5,-6
use trial and error and logic factor 120 120=2*2*2*3*5 how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120? obviously, the 5 has to stay since 2*5=10 which is out of range so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7 obviously, 7 cannot happen since it is prime 3 and 4 results in in 12, but 2*2*2*3=24 therfor answer is 4 and 6 they are all negative since negaive cancel except 1
