x(x-1)(x+1)(x-2)(x+3) = 144

x=3

two solutions

1. look at the factors of 144

2. subtract 144 and factor the left (to hunt down the other solutions)

method 1:

144's factors are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144

the numbers must have gaps of 1, 1, 1, 2 (x-2)(x-1)(x)(x+1)(x+3)

so it must be 3

method 2:

x(x-1)(x+1)(x-2)(x+3) = 144

x^5+x^4-7x^3-x^2+6x-144=0

the rational root theorem tells us that any rational roots must be factors of 144

we already found one up there with method 1

so x-3 must be a factor

(x-3)(x^4+4x^3+5x^2+14x+48)=0

lets look for other solutions with the rational root theorem

factors of 48 = +-(1, 2, 3, 4, 6, 8, 12, 16, 24, 48)

since all the coefficients are positive we know the other roots must be negative

i'm gonna cheat and use a graph, it turns out that none of the real numbers work

to the complex world it is then.

(someone help me on that part)