Results 1 to 4 of 4

Math Help - Is this correct factoring?

  1. #1
    Newbie
    Joined
    Feb 2010
    From
    Wilkes Barre, PA
    Posts
    15

    Is this correct factoring?

    Find all integers b for which a2 + ba 50 can be factored.
    Factors of 50 = (- 50, 1), (50 1), (- 25, 1), (25, 1), (- 10, 1), (10, -1), (- 5,1), (5,-1), (-2,1),(2,-1)
    (a 50)(a + 1) = a2 50a + a 50 = a2 49a 50
    (a + 50)(a 1 ) = a2 + 50a - a + 50 = a2 + 49a 50
    (a 25)(a + 1) = a2 + a - 25a 25 = a2 24a 25
    (a + 25)(a - 1) = a2 - a + 25a 25 = a2 + 24a 25
    (a 10)(a + 1) = a2 + a - 10a 10 = a2 9a 10
    (a + 10)(a - 1) = a2 - a + 10a 10 = a2 + 9a 10
    (a 5)(a + 1) = a2 + a - 5a 5 = a2 4a 5
    (a + 5)(a - 1) = a2 - a + 5a 5 = a2 + 4a 5
    (a 2)(a + 1) = a2 + a - 2a 2 = a2 a 2
    (a + 2)(a - 1) = a2 - a + 2a 2 = a2 + a 2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by mcruz65 View Post
    Find all integers b for which a2 + ba 50 can be factored.
    Factors of 50 = (- 50, 1), (50 1), (- 25, 1), (25, 1), (- 10, 1), (10, -1), (- 5,1), (5,-1), (-2,1),(2,-1)
    (a 50)(a + 1) = a2 50a + a 50 = a2 49a 50
    (a + 50)(a 1 ) = a2 + 50a - a + 50 = a2 + 49a 50
    (a 25)(a + 1) = a2 + a - 25a 25 = a2 24a 25
    (a + 25)(a - 1) = a2 - a + 25a 25 = a2 + 24a 25
    (a 10)(a + 1) = a2 + a - 10a 10 = a2 9a 10
    (a + 10)(a - 1) = a2 - a + 10a 10 = a2 + 9a 10
    (a 5)(a + 1) = a2 + a - 5a 5 = a2 4a 5
    (a + 5)(a - 1) = a2 - a + 5a 5 = a2 + 4a 5
    (a 2)(a + 1) = a2 + a - 2a 2 = a2 a 2
    (a + 2)(a - 1) = a2 - a + 2a 2 = a2 + a 2
    1. Explain the argument behind this

    2. Actually answer the question you have been asked

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    From
    Wilkes Barre, PA
    Posts
    15

    Am I approaching this problem correctly?

    The problem is the following:

    Find all integers b for which a^2 + ba – 50 can be factored.

    Am I solving this in the proper fashion?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by mcruz65 View Post
    The problem is the following:

    Find all integers b for which a^2 + ba – 50 can be factored.

    Am I solving this in the proper fashion?
    Factored over what? All quadratics can be factored over the complex numbers, so its unlikely to be that. Factored over the reals? Over the integers or rationals?

    The rational roots theorem tells you that if this quadratic has a factorisation with linear factors with only integer coefficients (which is equivalent to asking for factorisation over the rationals) then those factors correspond to roots:

    \pm 1, \ \pm 2,\ \pm 5,\ \pm 10,\ \pm 25,\pm 50

    this is the approach I believe you have adopted, and it is a workable approach for the problem. But you have to say what you are doing not just slap down an undigested bunch of expressions.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: July 29th 2011, 02:39 AM
  2. Can someone please tell me if this is correct.
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: December 15th 2009, 01:57 PM
  3. Is this correct?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 23rd 2009, 08:03 AM
  4. Replies: 2
    Last Post: August 22nd 2009, 11:57 AM
  5. Replies: 3
    Last Post: November 6th 2006, 12:02 AM

Search Tags


/mathhelpforum @mathhelpforum