Solving $x^2 - 16 > 0$ for $x$ This may be a very simple question, but I can't figure out the correct reason behind it.

If $x^2 - 16 >0$, which of the following must be true?
a. $4 < x$
b. $-4 > x > 4$
c. $-4 > x < 4$
d. $-4 < x < 4$

I know the answer but I didn't get how they figured out the direction.

Another approach would be to consider a factored form of the inequality:
$(x-4)(x+4)>0$
Now, for this to be positive, both factors have to share the same sign and so one could look at this as 2 cases:
1) $x-4>0$ and $x+4>0$ which would imply $x>4$
2) $x-4<0$ and $x+4<0$ which would imply $x<-4$
Then, one just combines these cases to get the proper answer.
