factoring ax2 + bx + c when a is not 1

Discussion in 'Secondary Education Archives' started by ms.jansen, Apr 21, 2007.

  1. ms.jansen

    ms.jansen Companion

    Joined:
    Jul 9, 2006
    Messages:
    205
    Likes Received:
    0

    Apr 21, 2007

    Fellow math people, I am stuck! My class is working on factoring ax2 + bx + c. No problems when a=1, but they are really struggling when it's not.

    So far I've just been having them make lists of factors for a and c then try a few of the combinations. From there they are supposed to see if b needs to be more or less and adjust which factors they used and in which order to try to find the correct factorization. They are getting frustrated that there is no "trick" to this and that sometimes there are 32 possible factorizations!

    It seems to me that the perfect square discriminant rule only applies when a=1 so now they have to try all the possibilities before they can say something is not factorable. Am I missing something? I have reread the section over and over and can't find anything new to tell them! Thanks!!!
     
  2.  
  3. Malcolm

    Malcolm Enthusiast

    Joined:
    Aug 25, 2005
    Messages:
    2,100
    Likes Received:
    1

    Apr 21, 2007

    The easiest way I have found to factor a quadratic where the coefficient of the first term is not 1 is "bottoms up". The kids like it better than any other way I have found.

    Example:

    Factor 2x^2 -5x -3

    Solve a diamond problem where the top number is the product of a and c and the bottom number is b:

    -6

    factor 1 factor 2

    -5

    Find the factors of -6 that add up to -5:

    -6

    -6 1

    -5

    Make binomial factors of the original quadratic from them:

    (x - 6)(x + 1)

    If a were 1, you would be done, but it is not.

    Now divide the constant terms in each binomial by a

    (x - 6/2)(x + 1/2)

    Reduce as far as possible

    (x - 3)(x + 1/2)

    If the constant term in either binomial is still a fraction, move the denominator in front of the x

    (x - 3)(2x + 1)

    You are done.
     
  4. Aliceacc

    Aliceacc Multitudinous

    Joined:
    Apr 12, 2006
    Messages:
    27,534
    Likes Received:
    6

    Apr 21, 2007

    I've seen people do it by Factoring by Grouping, but not often enough to explain it coherently.

    Sorry, I do trial and error:
    Factor the leading coefficient; split between the parentheses
    Factor the constant.
    Play with the combinations to find the one that gives the correct middle term.

    After enough examples, the kids get pretty fast at these. They still don't normally like them, but they can do them.
     
  5. ms.jansen

    ms.jansen Companion

    Joined:
    Jul 9, 2006
    Messages:
    205
    Likes Received:
    0

    Apr 21, 2007

    Thanks, Malcom and Alice. I've been using trial and error my whole life and I'm fast enough at it, but it's nice to know there is a method out there! I will present it to them but I'm sure some will stick with trial and error since they are getting faster now and a new way might confuse them. But next year I will use the "bottoms up" method. Thanks again, I knew I would learn something if I posted it here!
     

Share This Page

Members Online Now

  1. ready2learn,
  2. kellzy
Total: 411 (members: 7, guests: 352, robots: 52)
test