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Re: AC method of factoring polynomials
Posted:
Oct 26, 2013 10:25 AM


On Sunday, July 29, 2007 11:31:47 AM UTC7, Stephen J. Herschkorn 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. > >  > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhhattan
There is a "new and improved factoring AC Method", recently introduced (Google Search), that helps to simplify the AC factoring solving process. Back to the example: 18x^2 + 7x  30 = 0. Roots have opposite signs (Rule of Signs for Real Roots). Compose factor pairs of ac = 540 with all first numbers being negative. Since b is a small number, start the factor list from the middle of the list. Proceeding:...(10, 56)(15, 36)(20, 27) OK. This last sum is (20 + 27) = b. Then b1 = 20 and b2 = 27. Next, proceed solving by grouping as usual to get the 2 binomials: (9x + 10)(2x + 3) = 0. This "new and improved factoring AC Method" helps:  To know in advance the signs of the 2 real rood ( or +) for a better solving approach.  To reduce in half the number of factor pairs of (ac) to be composed, either in case roots have different signs or in case roots have same sign.  When a = 1, The 2 real roots can be immediately obtained from the factor pair list. There is no need neither to factor by grouping nor solving binomials.



