Results 1 to 10 of 10

Math Help - Factoring help

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    45

    Factoring help

    Hi, I need help with this question:

    Factor the expressions px^3 + (p-q)x^2 + (-2p-q)x + 2q and abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2

    I've tried a few different ways, but none of them work out. Thank you for any help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,667
    Thanks
    298
    Awards
    1
    Quote Originally Posted by starswept View Post
    Hi, I need help with this question:

    Factor the expressions px^3 + (p-q)x^2 + (-2p-q)x + 2q and abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2

    I've tried a few different ways, but none of them work out. Thank you for any help!
    There's either typos in this or there's something about the problem you aren't telling us. I don't believe these can be factored in general.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
     px^3 + (p-q)x^2 + (-2p-q)x + 2q
     px^3 + px^2 -2px - qx^2 - qx + 2q

     px(x^2 + x - 2) - q(x^2 - x - 2)

      px(x^2 + x - 2) - q(x^2 - x - 2)

      (px - q)(x + 2)(x - 1)
    Last edited by janvdl; July 7th 2007 at 02:54 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by topsquark View Post
    There's either typos in this or there's something about the problem you aren't telling us. I don't believe these can be factored in general.

    -Dan
    Oh yes they can
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2007
    Posts
    90
    px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px) says QuickMath

    We can make a list of the possible zeros, which include \pm 1, \pm 2, \pm q and check which yield a zero either through substitution or synthetic division.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2007
    Posts
    90
    You factored wrong on px(x^2 + x - 2) - q(x^2 - x + 2), Janvdl. The rest looks fine.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by rualin View Post
    px^3+(p-q)x^2+(-2p-q)x+2q = -(x-1)(x+2)(q-px) says QuickMath

    We can make a list of the possible zeros, which include \pm 1, \pm 2, \pm q and check which yield a zero either through substitution or synthetic division.
    i was just about to make that suggestion

    i noticed almost immediately that x - 1 was a factor by the factor theorem. performing long division twice will get you the answer you have

    coincidentally, (x - 1) is a factor of the second one as well

    janvdl's method is still easiest by far though. just make sure you factor correctly. factoring out a -q from the second set changes all the signs inside the bracket


    EDIT: Good Job Janvdl. You are the best mathematician in this post!
    Last edited by Jhevon; July 7th 2007 at 02:22 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jul 2007
    Posts
    90
    I also prefer that approach but when it seems difficult to factor, I find it better for me to use the factor theorem.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Jhevon View Post
    ..Good Job Janvdl. You are the best mathematician in this post!
    I'm flattered. Couldnt have done it without you and Rualin who showed me that mistake though. Thanks guys.
    Last edited by janvdl; July 7th 2007 at 02:43 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, starswept!

    abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2

    Expand: . abx^3 + ax^2 - 2bx^2 - abx^2 + 2bx - ax - 2x + 2

    Re-arrange: . \underbrace{abx^3 - abx^2} + \underbrace{ax^2 - ax} - \underbrace{2bx^2 + 2bx} - \underbrace{2x + 2}

    . . Factor: . abx^2(x-1) + ax(x-1) - 2bx(x-1) - 2(x-1)

    . . Factor: . (x-1)\,(\underbrace{abx^2 + ax} - \underbrace{2bx - 2})

    . . Factor: . (x-1)\,[ax(bx +1) - 2(bx+1)]

    . . Factor: . (x-1)\,(ax-2)\,(bx+1)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need help with factoring
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 7th 2010, 10:47 AM
  2. factoring help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 4th 2010, 07:44 PM
  3. Is this factoring or something?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 1st 2010, 06:54 PM
  4. Replies: 2
    Last Post: August 22nd 2009, 10:57 AM
  5. Replies: 3
    Last Post: November 5th 2006, 11:02 PM

Search Tags


/mathhelpforum @mathhelpforum