Multiplying Polynomials

Date: 07/10/2003 at 00:19:46
From: Jalise
Subject: Foiling

(x+2)(x+2)(x+2) 

How do you foil?

Date: 07/10/2003 at 12:21:22
From: Doctor Peterson
Subject: Re: Foiling

Hi, Jalise.

If you were told to use the FOIL method to do this, it won't work; it 
applies only to the product of two binomials. To do this, you will 
need a more general method for multiplying polynomials.

I'll demonstrate the process on a similar problem, so you can practice 
with your own. We'll expand (x+3)(x+3)(x+3).

First, we have to multiply (x+3)(x+3), which we can do just as in 
multiplying numbers: put one over the other, and multiply the top one 
by each term of the other:

         x + 3
  *      x + 3
  ------------
        3x + 9 <-- (x+3)3
  x^2 + 3x     <-- (x+3)x
  ------------
  x^2 + 6x + 9 <-- sum

So the first partial product, 3x + 9, is made by multiplying each term 
of x + 3 by 3, and the second, x^2 + 3x, by multiplying each term of 
x + 3 by x. We line them up so that like terms are in a column, and 
just add the columns.

Now we repeat, multiplying this by x + 3 again:

        x^2 +   6x  + 9
  *               x + 3
  ---------------------
        3x^2 + 18x + 27 <-- (x^2 + 6x + 9)3
  x^3 + 6x^2 +  9x      <-- (x^2 + 6x + 9)x
  ---------------------
  x^3 + 9x^2 + 27x + 27

Do you see how easy it is to do this way? But it's just another trick 
for multiplying every term of one polynomial by every term of the 
other.

Here is a page with some explanation of why we prefer not to teach 
FOIL, and links to alternatives:

   Why FOIL?
   http://mathforum.org/library/drmath/view/61769.html 

If you have any further questions, feel free to write back.

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