Reducing Polynomials to Lowest Terms

Date: 11/18/2002 at 13:39:04
From: M. W. Glugla
Subject: Reducing to the lowest terms

I am taking an algebra class, finally, for my first college degree - 
now that my kids are done it's my turn, but my algebra is more rusted 
than the Golden Gate Bridge.

How do I reduce the following:

( x^3+x ) (5x-5)
--------- ------
    5      x^3-x

The above has one bracket enclosing the entire fraction, not just the 
top line, but then the sum of each bracketed portion is multiplied by 
the other bracketed unit.

Thanks if you can shed light on this for me.

M. W. Glugla

Date: 11/18/2002 at 16:43:50
From: Doctor Ian
Subject: Re: Reducing to the lowest terms


Hi M. W., 

Let's take a look at the expression:

  (x^3 + x) (5x - 5)
  --------- --------
      5      x^3 - x

Whenever you see a polynomial, like 

  x^3 + x

the first thing you should be thinking about doing is factoring it.  

The simplest factoring technique is to use the distributive property.  
In 

  x^3 + x

both terms of the polynomial have x as a factor, so we can rewrite it
like this:

  x(x^2 + 1)

Let's do this to the other parts of the expression:

    (x^3 + x) (5x - 5)
    --------- --------
        5      x^3 - x

    x(x^2 + 1) 5(x - 1)
  = ---------- ----------
        5      x(x^2 - 1)

Now, note that on the right, we're multiplying by 5; and on the left,
we're dividing by 5. This is sort of like digging a hole and then
filling it up again. Instead of doing both operations, why don't we
skip them both, and call it even? 

    x(x^2 + 1)  (x - 1)
  = ---------- ----------
        1      x(x^2 - 1)

And why not do the same thing with x?

     (x^2 + 1)  (x - 1)
  = ---------- ----------
        1       (x^2 - 1)

Now what? The next thing to do is try to factor the expressions 
(x^2 + 1) and (x^2 - 1).

It turns out that the latter is an instance of a very common pattern,
called a 'difference of squares'. Whenever you see something like 

  this^2 - that^2

you should immediately think of 

  this^2 - that^2 = (this + that)(this - that)

We can use that pattern in the denominator on the right:

     (x^2 + 1)         (x - 1)
  = ---------- ----------------
        1       (x + 1)(x - 1)

Now we can do another cancellation, of (x-1):

     (x^2 + 1)     1
  = ---------- ---------
        1       (x + 1)

     (x^2 + 1)     
  = -----------
       (x + 1)


And that's as far as you can go without getting into complex numbers. 

The important thing to keep in mind is that you're always trying to
break things into factors, and then look for factors that cancel each
other out. It's really the same kind of thing that you're doing when
you simplify fractions, e.g., 

   28   15   2 * 2 * 7           3 * 5
   -- * -- = ----------------- * -----
   72   21   2 * 2 * 2 * 3 * 3   3 * 7

             2   2   3   7       5
           = - * - * - * - * --------- 
             2   2   3   7   2 * 3 * 3

                                 5
           = 1 * 1 * 1 * 1 * --------- 
                             2 * 3 * 3

Except now the factors can be variables (like 'x') or expressions
(like '(x-1)'). 

Does this make sense?  

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/