M18-16

x (x-1) (x+1) < 0

the solutions would require that of the given three factors (x, x-1, x+1) only if 1 or 3 of them are <0 then only the solution to x(x-1)(x+1) <0
i.e
x >0 and x+1>0 and x-1 <0
or x>0 and x-1>0 and x+1<0
or x+1>0 and x-1>0 and x< 0
or x<0 and x+1<0 and x-1<0


Now on solving

1: x >0 and x+1>0 and x-1 <0
we get x>0 and x>-1 and x<1
hence 0<x<1 is the solution range

2: x>0 and x-1>0 and x+1<0
we get x>0 and x>1 and x<-1
no possible solution range

3. x+1>0 and x-1>0 and x< 0
we get x>-1 and x>1 and x<0
no possible solution range

4. x<0 and x+1<0 and x-1<0
we get x<0 and x<-1 and x<1
hence the solution range is x<-1


therefore the solutions to x(x+1) (x-1) <0
are 0<x<1 or x<-1


another way to solve is to use the no. line
x(x-1)(x+1)<0

the solution of x(x-1) (x+1)=0 are x=-1,0,1

plotting these on no. line we get 4 regions

x<-1 , -1<x<0, 0<x<1 , and x>1

now for given exp: x(x-1) (x+1) we substitute for x values within the given range and check if it satisfies our criteria

if x<-1 exp <0

if -1<x<0 exp >0

if 0<x<1 exp <0

if x>1 exp >0


so for x(x-1) (x+1) <0
we need
x<-1 or 0<x<1