Rational functions w/ common factors don't simplify?

1. Jun 6, 2012

TheTinCook

Say we have a rational function P(x)=(x^2-3x-4)/(x-4)=[(x+1)(x-4)]/(x-4)

I'm a little confused as to why the (x-4) doesn't cancel out. It graphs the same as y=x+1 for x≠4. I feel like I'm missing something from the order of operations.

2. Jun 6, 2012

micromass

Staff Emeritus
You said it yourself, if $x\neq 4$, then

$$P(x)=x+1$$

But if x=4, then

$$P(x)\neq x+1$$

since the left-hand side is not defined, but the right-hand side is.

So the difference between P(x) and x+1 is that P(x) is not defined in 4, while x+1 is. So the two cannot be equal.

If your domain is $\mathbb{R}\setminus \{4\}$, then P(x)=x+1 does hold.

3. Jun 6, 2012

TheTinCook

Thanks!

I think the part I'm getting stuck on is why [(x+1)(x-4)]/(x-4)≠(x+1)

4. Jun 6, 2012

Number Nine

It has already been explained why: the left-hand side is undefined at x=4.

5. Jun 6, 2012

TheTinCook

I understand why the left hand side is undefined, I just don't get why it can't be reduced to the lowest terms in the first place.

6. Jun 6, 2012

micromass

Staff Emeritus
It can't be reduced since it's undefined. You can't do anything without undefined things.

7. Jun 6, 2012

DrewD

If the left hand side is undefined and the right hand side is defined, there is no equality. If there is no equality, then it can't be reduced. It doesn't work the other way. The rules of algebra stem from the facts about how numbers work, not the other way around.

If this were to come up in an application, and you knew you were never going near $x=4$ then you could reduce it. But in pure math, it does not work this way.

8. Jun 6, 2012

Hurkyl

Staff Emeritus
Eh? It works that way in "pure math" too; when x is restricted to being unequal to 4, both sides make sense and are equal.

9. Jun 6, 2012

TheTinCook

See, I get confused when I see things like this
from some college website I can't link to because I don't have enough posts yet.

This means if there is a real solution for R=0, then I can't just simplify it away?

So if the function was F(x)=[(X+1)(X^2+1)]/(X^2+1), I could factor out the (X^2+1) because it never = 0?

Thanks again for all your patient help. Sorry for being slow on the uptake.

10. Jun 6, 2012

Number Nine

You could factor that expression just fine, since [(X+1)(X^2+1)]/(X^2+1)] = X+1 for every real number.

11. Jun 6, 2012

micromass

Staff Emeritus
Yes,

$$\frac{(x+1)(x^2+1)}{x^2+1}=x+1$$

for all $x\in \mathbb{R}$. There are no undefined situations here for real x.

12. Jun 6, 2012

TheTinCook

Ah, I'm starting to get a glimmer of understanding. Thanks guys!

13. Jun 6, 2012

DrewD

You're right sorry. I was getting at the fact that often in physics and engineering the restriction isn't mentioned because of physical assumptions. It was an unnecessary comment.

14. Jun 6, 2012

Hurkyl

Staff Emeritus
You just had the misfortune of hitting a pet peeve of mine: when someone leaves an important assumption out of a problem, works on the problem, then suddenly uses that assumption to justify some step just before going on an aside to point out how important 'physical reasoning' is because you couldn't justify the step with mathematics.

While such reasoning is often justifiable, the aside really isn't, and I think I've most often seen trivial situations, like assuming $x \neq 4$ or that some function vanishes at infinity or some other such thing that is trivial to spell out.

And, for the record, 'physical reasoning' isn't the exclusive domain of physicists and engineers -- other disciplines reason in the same sort of fashion about their subject. Even mathematicians do, although I think they're more likely than most to review a completed argument to fix up the details.

15. Jun 7, 2012

HallsofIvy

Staff Emeritus
in other words, it is not a matter of "you can't cancel". What is true is that
$\frac{x^2- 3x- 4}{x- 4}= x+ 1$
as long as $x\ne 4$. Of course, they are not equal when x= 4 because then the right side is 5 and the left side is not defined.