Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: AC method of factoring polynomials
Posted:
Jul 30, 2007 12:20 AM


On Jul 29, 3:33 pm, quasi <qu...@null.set> wrote: > On Sun, 29 Jul 2007 14:31:47 0400, "Stephen J. Herschkorn" > > > > <sjhersc...@netscape.net> wrote: > >Summary: How well known and/or frequently taught is the AC method of > >factoring, sometimes called factoring by grouping. > > >Factoring of polynomials often seemed like an art to me. For example, > >consider > > >18x^2 + 7x  30. > > >I used to consider all possible pairs of factors of 18 and of 30 until > >I found the right coefficients. Considering placement of thes factors, > >that's 24 possible combinations, though with intuition (hence the art), > >I might be able to narrow down the search. > > > From a current client's textbook on College Algebra, I only recently > >learned a method the book calls "factoring by grouping." The client's > >professor calls it the "AC method," from consideration of polynomials of > >the type Ax^2 + Bx + C. Here's how it works in the above example: > > > Multiply the leading coeffiecient 18 = 2 x 3^2 and the constant > >term 30 = 2 x 3 x 5, getting 540 = 2^2 x 3^3 x 5. > > > Find a pair of factors of 540 such that their sum is the middle > >coefficient 7. That is equivalent to findiing factors of 540 whose > >difference is 7. Either by listing all the factors or by looking at the > >prime factorization, we find 20 = 2^2 x 5 and 27 = 3^3 as these > >factors. I prefer the prime factorization way, in which case I didn't > >even need the fact that the product was 540. > > > Rewrite the polynomial by splitting up the middle term: 18x^2 +27x  > >20x + 30. (20x + 27x will work as well.) > > > Factor by grouping: 9x(2x + 3)  10(2x + 3) = (9x 10) (2x + 3). > >Voil`a! (grave accent) > > > If no pair of factors of AC (the product of the leading coefficient > >and the constant term) sum to the middle coefficient B, then the > >polynomial is irreducible. > > >When A > 1, this approach seems in general a lot easier to me than > >searching pairs of factors of A and C individually. If you haven't > >seen this before, try it on some examples yourself, such as > > >6x^2 + 13x y + 6y^2 > >16a^4  24a^2 b + 9b^2 > >12x^2  29x + 15 > >6b^2 + 13b  28 > >10m^2 13m n  3n^2 > > >I don't think it is the case that I learned this method once long ago > >and subsequently forgot it, so I am surprised I never saw it before. > >How well known is this method? Is it taught much? I don't find it in > >my favorite College Algebra text (by C.H. Lehmann), and it doesn't show > >up in the first three pages from Googl(R)ing "polynomial factor." At > >least one of my more advanced clients had never seen it before either. > > It's sometimes called "the master product" method. > > In my opinion, it _shouldn't_ be taught  it's too artificial.
In practice, if you don't see how the polynomial can be factored, you should go to the quadratic formula. This is pretty much what I say about symmetry in Calculus problems: If you can see right off that there is symmetry, go ahead and use it (if you can), but don't spend time looking for something that might not even be there (especially on a test).
 Christopher Heckman
> Sure it works, and it's not hard (for us) to prove that it works, but > for the student of elementary algebra, it's just some kind of > meaningless magic. > > Moreover, it obscures the much simpler, much more basic idea that the > leading coefficients of the factors must produce the leading > coefficient of the product, and similarly for the constant terms. > > Also, students are not in the _business_ of factoring nonmonic > quadratics. It's not like there are going to be that many of them. > Much better if they simply work through factor tables for both the > leading and trailing coefficients, since that method that actually > makes sense, even if it's a little longer. > > Besides, at the next level of algebra (intermediate algebra or > precalculus), you can factor a nonmonic quadratic by first completing > the square  thus, no trial and error at all. Alternatively, you can > use the quadratic formula to find the roots, and from the roots, you > can deduce the factors. > > The students have enough to learn. No need to burden them with obscure > methods that get obsoleted at the next level. > > quasi



