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

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quasi

Posts: 10,226
Registered: 7/15/05
Re: AC method of factoring polynomials
Posted: Jul 29, 2007 4:33 PM
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On Sun, 29 Jul 2007 14:31:47 -0400, "Stephen J. Herschkorn"
<sjherschko@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.

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 non-monic
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 non-monic 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



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