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Why Does the Algorithm for Polynomial Long Division Ignore Smaller Terms?

Date: 03/24/2011 at 17:52:21
From: Cody
Subject: Proof/Explanation why long division works with polynomials.

I've read this post on long division by numbers:

  http://mathforum.org/library/drmath/view/70275.html 

Despite a lot of searching, I want to see more of why the vertical
manipulation works. Here is an example of what I'm looking for.

Example

   (2x^2 + 7x + 5) divided by (2x + 5).

This is easily divided out by long division to get (x + 1).

My question is about the division PROCESS to get the quotient (x + 1).

Why/how does the process work? What's the proof to the process that makes
it start out with 2x^2/2x = x, then multiply back through the binomial
divisor, then subtract and repeat with a new line?

How can we show/prove that this "algorithm" works for all division for
polynomials with divisor degree less than that of the dividend?

Basically, I want a simple explanation to the question, "Why do we divide
by the first term and continue while ignoring the other terms until the
multiplication?"

For my example, I purposely chose a polynomial for the divisor that is not
a monic. Are there two cases?



Date: 03/24/2011 at 23:30:24
From: Doctor Peterson
Subject: Re: Proof/Explanation why long division works with polynomials.

Hi, Cody.

Any polynomial works the same way -- and for the same reasons.

Let's examine division in light of what it is supposed to do. Division
means finding the other factor to get a given product; for example, with
numbers, dividing 7 by 3 means finding what number q we can multiply by 3
to get (close to) 7:

   7 / 3 = 2 R 1

means that

   7 = 3 * 2 + 1

So we can understand a division by seeing how it arrives at the required
other factor.

I'll first multiply two simple polynomials and add a remainder, in order
to create a division problem the answer to which we know ahead of time:

          3x + 1 <-- factor (divisor)
   *      2x - 3 <-- factor (quotient)
   -------------
         -9x - 3 <-- partial product (-3)(3x + 1)
   6x^2 + 2x     <-- partial product (2x)(3x + 1)
   -------------
   6x^2 - 7x - 3
             + 4 <-- addend (remainder)
   -------------
   6x^2 - 7x + 1 <-- product (dividend)

So if we divide 6x^2 - 7x + 1 by 3x + 1, we should get a quotient of
2x - 3, with a remainder of 4.

Let's start. Reversing the process of multiplication (in which we
repeatedly ADDED multiples of 3x + 1 by single terms), we want to SUBTRACT
appropriate multiples of 3x + 1 by single terms, building the quotient
term by term as we dismantle the dividend. Each subtraction should
eliminate a term of the dividend, so that we will reduce its degree at
each step.

What multiple of 3x + 1 can we subtract from 6x^2 - 7x + 1 in order to
eliminate the leading term? Well, in order to have ...

   (3x + 1)(__ + ...) = 6x^2 + ...

... the blank has to be 2x. Note that since we are focused on eliminating
the leading term of the dividend, and in multiplication that term is
formed by multiplying the leading terms of the factors, all we have to
look at in making this decision are the leading terms 6x^2 and 3x. That
answers your main question.

So the first term of our quotient has to be 2x. That gives us

  (3x + 1)(2x + __) = (6x^2 + 2x) + (3x + 1)(__)

Now we can subtract the product we have so far, 6x^2 + 2x, from 
6x^2 - 7x + 1 in order to see what more we need:

      6x^2 - 7x + 1
   - (6x^2 + 2x    )
   -----------------
            -9x + 1

Now for our next term to eliminate the new leading term, we need

   (3x + 1)(__) = -9x + ...

What can fill in the blank to get this leading term? -3. Let's try it:

   (3x + 1)(-3) = -9x - 3

Look back at the multiplication I did above. In doing our divisions, we've
been working backward, discovering the partial products needed to add up
to the dividend. We first found the partial product 6x^2 + 2x that
accounts for the 6x^2 in the dividend; we subtracted that off to see that
the next partial product had to account for a -9x, and we saw that it was
-9x - 3. Subtracting off this last partial product, we see that we still
have to add 4 to get the dividend.

The work we just did so laboriously is the following:

                   2x - 3
          ----------------
   3x + 1 ) 6x^2 - 7x + 1
          -(6x^2 + 2x    ) <-- (2x)(3x + 1)
            -------------
                  -9x + 1
                -(-9x - 3) <-- (-3)(3x + 1)
                  -------
                        4

What this means, expanded, is just a repeated subtraction:

   (6x^2 - 7x + 1) - (6x^2 + 2x)  - (-9x - 3)    = 4
 
   (6x^2 - 7x + 1) - (2x)(3x + 1) - (-3)(3x + 1) = 4

   (6x^2 - 7x + 1) - (2x - 3)(3x + 1)            = 4

   (6x^2 - 7x + 1)                               = (2x - 3)(3x + 1) + 4

And this is what we were trying to accomplish.

Does that help? Again, the reason we look at the leading terms of the
divisor and (current) dividend at each step is so that what we subtract
will reduce the degree of the dividend, leaving us eventually with only a
remainder (with degree less than the divisor).

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra

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