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/
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