Simple Algrebra: Factoring quadrinomials

Discussion in 'Community Discussion' started by floriflee, May 20, 2007.

  1. floriflee macrumors 68030

    floriflee

    Joined:
    Dec 21, 2004
    #1
    I feel silly for having to post this, but for the life of me I can't remember how to factor quadrinomials. My mother is taking an algebra class and was asking for my help. Shamefully, I couldn't figure out how to get the answer. I'd like to blame it on the fact that I haven't done this kind of math in well over 10 years, and that it's so late so my brain isn't working. So here I am posting to find out how to solve this problem because it's going to drive me crazy until I figure out what it is that I'm missing:

    x^3 + 4x^2+x -6

    The answer is (x-1)(x+2)(x+3), but how do you get to that?
    I've tried various forms of grouping, but so far nothing has come up as a GCF for both groups. What am I forgetting?
     
  2. siurpeeman macrumors 603

    siurpeeman

    Joined:
    Dec 2, 2006
    Location:
    the OC
    #2
    you could try synthetic division. looking at the coefficient of x^3 and the last term, the only possible factors are ±1, ±2, ±3 and ±6**. once you've found one factor, the other two shouldn't be hard.

    **this is because of the rational root theorem.

    *edit*
    you can make your guessing and checking a little easier by knowing there's only one positive root (descarte's rule of signs), as there is only one sign change. testing +1 will give you your first root on the first go.
     
  3. swiftaw macrumors 603

    swiftaw

    Joined:
    Jan 31, 2005
    Location:
    Omaha, NE, USA
    #3
    Consider a factored cubic (x-a)(x-b)(x-c)

    Expanding this you get x^3 - (a+b+c)x^2 + (ab + bc + ac)x - abc

    thus, if you are trying to find the factors of x^3 + 4x^2+x -6 you know that

    (a+b+c) = -4
    ab+bc+ac = 1
    abc = 6

    3 equations for 3 unknowns can be solved for the desired result.

    Of course, another method is trial and error to find one of the roots, then that will leave you with a quadratic which is easy to factorize.
     
  4. floriflee thread starter macrumors 68030

    floriflee

    Joined:
    Dec 21, 2004
    #4
    Ah, yes... synthetic division. I'd completely forgotten about that. Hopefully, they've covered that in her class already--I can't remember which order I learned what in in my classes....

    Thanks, again!
     
  5. ethernet76 macrumors 6502a

    Joined:
    Jul 15, 2003
    #5
    You also should be able to graph the equation to find the roots.

    Anywhere the graph crosses the line is a factor.

    Grapher shows it crosses it at -3, -2, and 1.

    Taking the negative of each results in (x+3), (X+2) and (x-1)
     
  6. floriflee thread starter macrumors 68030

    floriflee

    Joined:
    Dec 21, 2004
  7. siurpeeman macrumors 603

    siurpeeman

    Joined:
    Dec 2, 2006
    Location:
    the OC
    #7
    no offense but i think you made the problem worse by doing what you did. just because you get three equations with three unknowns doesn't mean the solution is going to be a walk in the park. normally, you solve systems of equations by isolating and eliminating one variable from two equations. then you do the same with two other equations to get two equations with two unknowns. this simply isn't practical, especially with your second and third equations. i mean, could you solve it?

    this is the best i could do:
    ab + bc + ac = 1
    a + b + c = -4
    abc = 6

    b (a + c) + ac = 1
    -b (b + 4) + ac = 1
    -b (b + 4) + 6/b = 1
    -b ^2 - 4b + 6/b = 1
    b^2 + 4b + 1 - 6/b = 0
    b^3 + 4b^2 + b - 6 = 0

    working with all three equations got me back to the original problem, which i would solve using synthetic division.
     
  8. swiftaw macrumors 603

    swiftaw

    Joined:
    Jan 31, 2005
    Location:
    Omaha, NE, USA
    #8
    My Bad, they are solvable, however to solve them means solving a cubic, which was the original problem so in this case not worth the effort. I must admit I did set up the equations without trying to solve them, all I know is that they could be solved - but it makes no sense to introduce a method which is just a complex as the original question. Oh well, serve me right for giving a rushed solution and not finishing my work. :)
     
  9. cruxed macrumors regular

    cruxed

    Joined:
    Aug 28, 2005
    Location:
    Hong Kong
    #9
    I typed up the working pretty briefly in Microsoft Word.
    The good thing about long division is that it can be used not only for quadronomials, but probably ever other polynomial.

    Hope it helps :)
     

    Attached Files:

    • Math.doc
      File size:
      33.5 KB
      Views:
      104
  10. polevault139 macrumors 6502

    Joined:
    Sep 24, 2006
    Location:
    Illinois
    #10
    It's funny, since I have been taking claculus I sometimes forget some of the easier algebra. Luckily the AP Test is over and I can start to regain my sanity.
     
  11. notjustjay macrumors 603

    notjustjay

    Joined:
    Sep 19, 2003
    Location:
    Canada, eh?
    #11
    No kidding! I have TWO engineering degrees, and I've done YEARS of calculus (and other courses which were really just calculus in disguise, like physics) and yet I can't even remember any of this stuff anymore. :(
     
  12. siurpeeman macrumors 603

    siurpeeman

    Joined:
    Dec 2, 2006
    Location:
    the OC
  13. jonasfan393 macrumors newbie

    Joined:
    Jun 24, 2009
    #13
    Help with this problems!

    My problem has two different variables and I can't figure it out. Here it is: 15a^2-3x-3+7a
     

Share This Page