Factoring by Grouping to Solve a Polynomial EquationDate: 05/26/2005 at 21:06:47 From: Kate Subject: re: Factoring and Solving Polynomial Equations How do you factor polynomials when there is an x cubed in the equation and it doesn't fit into a perfect square form? For example: x^3 + 3x^2 - x - 3 = 0 I think that you would first have to add 3 to both sides: x^3 + 3x^2 - x = 3 Then you would be able to take an x out of the equation: x(x^2 + 3x - 1) = 3 The problem is that there are no factors of -1 which will work together to give you a positive 3 in the middle. Would you then have to use the quadratic formula to solve it? That seems way too messy. Date: 05/26/2005 at 23:11:14 From: Doctor Peterson Subject: re: Factoring and Solving Polynomial Equations Hi, Kate. To solve an equation by factoring, the other side has to be zero, so you don't want to change the right side by adding 3. In factoring, you are paying attention only to the one expression you are factoring, even if it is part of an equation. In general, a quadrinomial (four-term polynomial) is very hard to factor. So if you are given one, you can expect it to be one you can do using one of the few available techniques. They are: 1. Factor out a common factor (as you tried to do with the x) 2. Factor by grouping (see below) 3. Factor three terms as a trinomial, and then combine that with the other term as a difference of squares The last possibility is a rarity. Your problem can be done by grouping. Then idea is to group the terms into two pairs, and factor out the greatest common factor in EACH PAIR. If it works right, the resulting two factored expressions will then have a common factor that you can pull out. Here's an example: 2x^3 - 3x^2 - 4x + 6 We group the first two terms and the last two terms, and factor out the GCF from each pair: (2x^3 - 3x^2) - (4x - 6) x^2(2x - 3) - 2(2x - 3) Now we see that the two "terms" both have a factor of 2x-3, so we factor that out: (2x - 3)(x^2 - 2) If we were lucky, we could now factor that second term (if it were a difference of squares); since we can't, we're done. Can you do that with your problem? If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/