Simplifying (x^2 - 9)/(x^2 - 2x - 3) has me stumped

ribald

New member
Joined
Jun 21, 2018
Messages
2
The original expression is (x^2 - 9)/(x^2 - 2x - 3)

This can be simplified by factoring to [(x + 3)(x - 3)]/[(x + 1)(x - 3)]

That can further be simplified by cancelling out (x - 3) on top and bottom: (x + 3) / (x + 1)

So, (x^2 - 9)/(x^2 - 2x - 3) = (x + 3) / (x + 1)

But when I plug in numbers (like x = 3), that isn't true.

What am I missing?

PS - this problem was actually a limit problem, but I don't think that should matter? It said "with limits we can cancel the x-3 because x approaches 3, hence x-3 gets close to zero, but not equal to zero". Shouldn't you always be able to cancel an (x-3) that is on the top and bottom of a fraction like that?
 
The original expression is (x^2 - 9)/(x^2 - 2x - 3)

This can be simplified by factoring to [(x + 3)(x - 3)]/[(x + 1)(x - 3)]

That can further be simplified by cancelling out (x - 3) on top and bottom: (x + 3) / (x + 1)

So, (x^2 - 9)/(x^2 - 2x - 3) = (x + 3) / (x + 1)

But when I plug in numbers (like x = 3), that isn't true.

What am I missing?

PS - this problem was actually a limit problem, but I don't think that should matter? It said "with limits we can cancel the x-3 because x approaches 3, hence x-3 gets close to zero, but not equal to zero". Shouldn't you always be able to cancel an (x-3) that is on the top and bottom of a fraction like that?

"...cancelling out (x - 3)"

You have just tripped over the reason why this language is inadequate, confusing, and, in my opinion, forgettable.

\(\displaystyle \dfrac{x-3}{x-3}\) is not NOTHING. You can't just discard it.

\(\displaystyle \dfrac{x-3}{x-3} = 1\) IF \(\displaystyle x \ne 3\)

What you should have done, in addition to forgetting that silly "cancel' word, is this:

\(\displaystyle \dfrac{(x-3)(x-3)}{(x+1)(x-3)} = \dfrac{x-3}{x+1}\cdot\dfrac{x-3}{x-3}\)

Then, you state this. \(\displaystyle \dfrac{x-3}{x+1}\cdot\dfrac{x-3}{x-3} = \dfrac{x-3}{x+1}\cdot 1 = \dfrac{x-3}{x+1}\) EXCEPT where \(\displaystyle x = 3\)

After saying that, are you still tempted to try x = 3? The original expression and the final expression are equivalent EXCEPT for x = 3. No more "cancelling". Resist the cursed beast!

"Simplify" isn't magic. It has bounds and limitations. No matter what you do, you still have to remember where you started.
 
PS - this problem was actually a limit problem, but I don't think that should matter? It said "with limits we can cancel the x-3 because x approaches 3, hence x-3 gets close to zero, but not equal to zero". Shouldn't you always be able to cancel an (x-3) that is on the top and bottom of a fraction like that?

If it is a limit problem, why are you substituting? Don't do that. Consider the limit, the approach. Sneak up on it.

Personally, I would almost never consider substitution, but I am often reminded that one can use substitution AFTER you prove things continuous. This one is NOT continuous at x = 3. Thus, x = 3 messes you up and provides NO information except that you have some misconceptions about "simplify", "cancel", "approach", and "limit". Feel free to get the misconceptions out of your system.

Note: After you "simplified", you created an ALMOST IDENTICAL structure that is continuous at x = 3. Too bad x = 3 is not in the Domain. This should fascinate you.
Alternate Note: After you "simplified", you created a structure that was IDENTICAL to the original only because you remembered to discard x = 3. This should fascinate you.
 
Those responses pretty much clear things up for me. Much appreciated.

I'm studying right now for the PCAT and having to re-learn many of the old lessons from high school math. It's kinda fun indeed.
 
Top