tag:blogger.com,1999:blog-8399389267815059112.post1814813246532390365..comments2023-09-10T05:21:25.471-05:00Comments on Algebra's Friend: Factoring ... which method works best?Algebra's Friendhttp://www.blogger.com/profile/04729315514507170702noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-8399389267815059112.post-2885078950863305762013-11-11T14:22:05.865-06:002013-11-11T14:22:05.865-06:00Not anymore. I used to place value on just sort of...Not anymore. I used to place value on just sort of sifting through options in your head - but I feel that a systematic way (especially if it reinforces something else that's important) is a better approach.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8399389267815059112.post-34713020495106926732013-11-11T06:58:16.531-06:002013-11-11T06:58:16.531-06:00The method you describe is what we call factoring ...The method you describe is what we call factoring by grouping. Do you teach any other method in your classes?Algebra's Friendhttps://www.blogger.com/profile/04729315514507170702noreply@blogger.comtag:blogger.com,1999:blog-8399389267815059112.post-65272776490840164492013-11-11T04:52:03.176-06:002013-11-11T04:52:03.176-06:00I cannot get your link to work (the I read about i...I cannot get your link to work (the I read about it here link) but I suspect that I know and use what you call the AC method. I tell my students that quadratics of the form ax^2 + bx + c yield to this technique<br />1) Find the product ac<br />2) Find factors of ac that add to b - let's call them p and q<br />3) Rewrite ax^2 + bx + c as ax^2 + px + qx + c<br />4) Since p and q are factors of ac we know for sure that ax^2 + px has a common factor - let's call it rx. The remaining factor is of special interest.<br />5) Since p and q are factors of ac we know that qx + c will have a common factor - let's call is sx. The remaining factor when we pull out sx is exactly the same as the remaining factor from step 4. We know have decomposed ax^2 + bx + c<br /><br /><br />The reason I especially like this technique is that it plays into my disdain for FOIL-ing and shows how this process is simply the opposite of distributing to multiply binomials. <br />Hope this made sense!<br />Anonymousnoreply@blogger.com