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Factorize: (n^2 + 2n +3 ) / (2n^3 + 5n^2 + 8n +3)

  1. Sep 5, 2006 #1
    I have a really bad time thinking on this:

    Factorize:

    (n^2 + 2n +3 ) / (2n^3 + 5n^2 + 8n +3) such that we end up with 1/(2n +1).

    I might be forgetting how to complete the square or the cube or something but I can not find a way to reduce it.

    Any advice?
     
  2. jcsd
  3. Sep 5, 2006 #2

    berkeman

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    Staff: Mentor

    You should be able to factor the numerator. Do you know how to set up the two simultaneous equations to figure out the a and b in (n+a)(n+b) = n^2+[a+b]n+[a*b] ?
     
  4. Sep 5, 2006 #3
    Answer

    Say for n^2 + 2n + 3 we can write it as (n+1)^2 + 2, so we have that (n+1)(n+1) + 2 = n^2 + 2n + 3
     
  5. Sep 5, 2006 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are told that you must wind up with an extra factor of 2x+1 in the denominator. Is it that hard to divide n2 + 2n +3 and 2n3 + 5n2 + 8n +3 by 2n +1?
     
  6. Sep 5, 2006 #5

    berkeman

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    Staff: Mentor

    LOL, that's so much better a way to start. I suggested the brute strength way to start, but spaced the alternate constant form. Good stuff jetoso.
     
  7. Sep 6, 2006 #6
    I got it

    Well, I think I did not pay too much attention to this problem, it is very straight forward by the way, look:

    2n^3 + 5n^2 + 8n + 3 = (2n + 1) (n^2 + 2n + 3)

    Thus, since the numerator is n^2 + 2n + 3, then it follows that:

    (n^2 + 2n + 3)/(2n + 1) (n^2 + 2n + 3) = 1/(2n + 1)


    Sorry to bother you about this.
    Thanks!
     
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