Factoring a Difference of Squares with Terms That Are Not MonomialsDate: 10/26/2008 at 21:55:53 From: Aaron Subject: Factoring difference of squares that are not monomials I need to know how to do this, and I can't do it..like here's an example if I'm not making sense. X power 4 minus (2X - 1) squared. That's an example out of my text book but I don't understand there method of solving. Could you please give me another example or just explain it? Date: 10/26/2008 at 22:58:20 From: Doctor Peterson Subject: Re: Factoring difference of squares that are not monomials Hi, Aaron. First, I assume you understand the general rule: any difference of squares can be factored as a sum times a difference: (a)^2 - (b)^2 = (a + b)(a - b) The "a" and "b" in this formula can be ANYTHING--a single variable, a monomial, or a more complicated expression. The key is to identify what "a" and "b" are in your expression and write it out to make them clear before actually factoring. In your example, you'd have (x^2)^2 - (2x - 1)^2 so that "a" would be represented by x^2 and "b" would be represented by (2x - 1). Here's an example that's even more complicated than yours: 9(x-1)^2 - 4(x+1)^2 First, we see that each term is the square of something; I'll write them as squares to make that plain: [3(x-1)]^2 - [2(x+1)]^2 That is, inside the [] I wrote the square root of each part of a term, showing what had to be squared to get it: the 9 is 3 squared, and the (x-1)^2 is already written that way. Now, in my mind (or maybe on paper, if I'm showing it to someone) I'll label "a" (the first thing being squared) and "b" (the second thing being squared): [3(x-1)]^2 - [2(x+1)]^2 \____/ \____/ a b Now, the formula says that the first factor will be a+b, so I write the sum of those two expressions; and for the second factor I write the difference of the two: (3(x-1) + 2(x+1))(3(x-1) - 2(x+1)) \____/ \____/ \____/ \____/ ( a + b )( a - b ) Now we've factored the expression; but we should do some cleaning up before we say we're done. I'll simplify INSIDE each factor by distributing and combining like terms. But I'll leave each factor intact, since that was my purpose. (3(x-1) + 2(x+1))(3(x-1) - 2(x+1)) (3x - 3 + 2x + 2)(3x - 3 - 2x - 2) <-- watch that last sign! (5x - 1)(x - 5) And we're done! To check, you might expand both the original ugly expression, and this factored one (distributing and combining terms), and see that they both represent the same polynomial. If you have any further questions, feel free to write back. The best thing would be to show your work on a specific problem so I can see how close you are to getting it right. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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