Question

In an inequality such as (x+1)(x+2)(x+3)>0, which of the following is not a reason why setting each factor greater than zero and solving for x does not produce all of the solutions?

A) It is possible that (x+1)<0, (x+2)<0, and (x+3)<0.
B) It is possible that (x+1)<0, (x+2) <0, and (x+3)>0
C) It is possible that (x+1)>0, (x+2)<0, and (x+3)<0
D) It is possible that (x+1)<0, (x+2)>0, and (x+3)<0

Answer 1

confusing question-- too many negatives!

B,C, and D are all ways the LHS can be positive, so (if I understand the question) the answer must be A; A is the only one that would make the LHS negative (three negative factors)

Answer 2

i got it! wattayaknow