(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive 1 case) x+12<0, and x+3>0, x<-12, and x>-3 (-∞, -12) and(-3,∞) gives empty set
or second case) x+12>0 and x+3<0 x>-12 and x<-3 (-12,∞) and (-∞,-3) they are crossing , so (-12, -3) is a solution of this inequality