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.
 A: Your inequality $$x^2-16 > 0$$ factors as $$(x-4)(x+4) > 0.$$  Since the product of $(x-4)$ and $(x+4)$ is positive, they must have the same sign.


*

*Suppose $x-4$ and $x+4$ are both positive.  But $x+4$ is bigger than $x-4$, so it is enough for $x-4$ to be positive; if it is, then the other one will be too.  $x-4$ is positive whenever $x>4$.  

*Or both  $x-4$ and $x+4$  could be negative.  But $x+4$ is bigger than $x-4$, so if $x+4$ is negative, $x-4$ is also.  And $x+4$ is negative whenever $x<-4$.
So the solution is that either $x<-4$ or $x>4$.
Looking at the multiple-choice, I have no idea which to choose; they all seem wrong.
Whoever wrote the question should get a kick in the pants for writing the absurd “$-4 > x < 4$” and the patently false “$-4 > x > 4$”.
A: Since $(-4)^2=4^2=16,$ and for $x^2-16>0$ to be true, $x$ has to be strictly greater than $4$ and strictly less than $-4,$ then from this I think you can tell which answer must be correct
A: 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.
