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Topic: AC method of factoring polynomials
Replies: 5   Last Post: Oct 26, 2013 10:25 AM

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 nghi_nguyen@sbcglobal.net Posts: 1 Registered: 10/26/13
Re: AC method of factoring polynomials
Posted: Oct 26, 2013 10:25 AM

On Sunday, July 29, 2007 11:31:47 AM UTC-7, 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.

Date Subject Author
7/29/07 Stephen J. Herschkorn
7/29/07 quasi
7/30/07 Proginoskes
7/29/07 Inansawa Khin
7/31/07 Dave L. Renfro
10/26/13 nghi_nguyen@sbcglobal.net