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### Simplifiying Rational Expressions

```Date: 11/29/2004 at 11:16:29
From: marielys
Subject: how do I solve rational expressions

How do I solve something like this:

x + 2          x
-------- - -------------
x^2 - 36   x^2 + 9x + 18

We've been working on factoring, and I was doing fine until this
section came along.

```

```
Date: 11/29/2004 at 14:28:08
From: Doctor Ian
Subject: Re: how do I solve rational expressions

Mi Marielys,

Let's look at an example using simpler expressions.  Suppose I have

3    2
-- + -- = ?
2x   3y

and I want to simplify this.  I need a common denominator, right?  To
get a common denominator, I can multiply each fraction by ()/(), where
() is the denominator of the other fraction:

3(3y)    2(2x)   3(3y) + 2(2x)                      9y + 4x
------ + ------ = ------------- to give an answer of -------
2x(3y)   3y(2x)     (2x)(3y)                           6xy

Does this make sense?  Your problem is the same, but with more
complicated expressions.

So in your first problem, you could do this:

(x+2)(x^2+9x+18)           (x)(x^2-36)
------------------- - -------------------
(x^2-36)(x^2+9x+18)   (x^2+9x+18)(x^2-36)

Now, before you have a heart attack at the prospect of having to
simplify this monster, do you see why it's really the same concept
you've always used to add fractions, finding a common denominator and
renaming each fraction so it has that denominator?  Once you have that
down, the rest is just a matter of dealing with increased complexity.

Okay, one way to deal with increased complexity is to write things in
terms of factors (which is why you are doing this after learning
factoring), rather than sums.  For example, note that

x^2 - 36

is a difference of squares, which can be factored thus:

x^2 - 36 = x^2 - 6^2

= (x + 6)(x - 6)

So we can rewrite our expression as

(x+2)(x^2+9x+18)           (x)(x+6)(x-6)
--------------------- - ---------------------
(x+6)(x-6)(x^2+9x+18)   (x^2+9x+18)(x+6)(x-6)

It's probably not a coincidence that

x^2 + 9x + 18

can be factored too:

x^2 + 9x + 18 = (x + 3)(x + 6)

So we can make this substitution, too:

(x+2)(x+3)(x+6)          (x)(x+6)(x-6)
-------------------- - --------------------
(x+6)(x-6)(x+3)(x+6)   (x+3)(x+6)(x+6)(x-6)

There's no more factoring, so we can go ahead and do the addition:

(x+2)(x+3)(x+6) - (x)(x+6)(x-6)
-------------------------------
(x+6)(x-6)(x+3)(x+6)

Now, it's tempting to just go ahead and expand all the factors out
again, and do the addition; but one big benefit of having done the
factoring is that we can use the distributive property to get some
cancellations going:

(x+2)(x+3)(x+6) - (x)(x+6)(x-6)
-------------------------------
(x+6)(x-6)(x+3)(x+6)

(x+6)[(x+2)(x+3) - (x)(x-6)]
= ----------------------------
(x+6)(x-6)(x+3)(x+6)

(x+2)(x+3) - (x)(x-6)
= ---------------------
(x-6)(x+3)(x+6)

That may not seem like a big help, but it means that if we _do_ want
to expand the numerator, we can deal with a quadratic expression
instead of a cubic.  I don't know about you, but I see that as a win.

Okay, expanding, we get

(x^2+5x+6) - (x^2-6x)
= ---------------------
(x-6)(x+3)(x+6)

x^2 + 5x + 6 - x^2 + 6x
= -----------------------
(x-6)(x+3)(x+6)

5x + 6 + 6x
= ---------------
(x-6)(x+3)(x+6)

11x + 6
= ---------------
(x-6)(x+3)(x+6)

Now, that was kind of a mess, but not too bad.  What I sort of
expected to happen was to end up with something factorable in the
numerator, e.g.,

(x-6)*(...)
-------------
(x-6)(x+3)(x+6)

so that another of the binomials in the denominator would cancel.

Now, note that I could have factored even sooner, and that would have

(x+2)       (x)
-------- - -----------
(x^2-36)   (x^2+9x+18)

(x+2)        (x)
= ---------- - ----------
(x+6)(x-6)   (x+6)(x+3)

Now, instead of just taking the 'obvious common denominator', I can
find the _least_ common denominator:

(x+2)(x+3)          (x)(x-6)
= --------------- - ---------------
(x+6)(x-6)(x+3)   (x+6)(x+3)(x-6)

(x+2)(x+3) - (x)(x-6)
= ---------------------
(x+6)(x-6)(x+3)

which gets us to the same place before, but with a little less pain.

Anyway, the basic idea is always the same:  Factor as soon as
possible, find a common denominator, do the addition, expand the
numerator, factor the numerator again if possible, and look for
opportunities to cancel binomial (or other) factors.

As I noted earlier, it's really the same sort of thing you've already
been doing forever.  The hardest thing is to avoid being intimidated.

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Factoring Expressions

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