(x + 3)(x - 3) > 0
=> x + 3 > 0 or x - 3 > 0
=> x > -3 or x > 3
now, with inequalities, things sometimes work out weird, and in evaluating algebraically, you might get the signs turned the wrong way. so here we test the answers. we have two values to work around, x = -3 and x = 3. x cannot be equal to either of these values since that would give zero, and we must be greater than zero. so now, let's check to the left and right of these numbers.
if x < - 3, is (x + 3)(x - 3) > 0 ? .........YES! so this is one solution
if -3 < x < 3, is (x + 3)(x - 3) > 0 ? .....NO! this isn't a solution
if x > 3, is (x + 3)(x - 3) > 0 ? .............YES! this is another solution
so finally, our solution is, x < -3 or x > 3
it looks long and tedious because i typed out the steps, but most of what i typed you should be able to do in your head somewhat easily, if not, very easily.
2. x^2 -2x-15 < 0
Now continue the way I did in the previous question
this is a nice one, there is no constant term to worry about, so we simply get everything on one side and factor out the common terms
................put everything on one side
............now factor out the common term
You finish this one. remember what i said about two numbers being multiplied give zero