What are 3 consecutive integers whose product is - 120

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

the numbers are - 4,-5,-6