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Simplify using Factoring after Quotient Rule

  1. Aug 26, 2010 #1
    I am taking an online Introductory Calculus course. I have a decent understanding thus far, however, the problem I'm working on gets somewhat messy and I am having a difficult time simplyifing the answer.

    f"(x) = (x^2 + 9)^2 (-2x) - [(9 - x^2)(2)(x^2 + 9)(2x)]/(x^2 + 9)^4

    the solutions manual I have been given gives the following simplified answer:

    = (2x)(x^2 + 9) [-(x^2 + 9) -2(9 - x^2)]/(x^2 +9)^4

    = 2x(x^2 - 27)/(x^2 + 9)^3

    I am having a difficult time understanding the factoring involved in the algebraic simplification. I understand how to factor by grouping, trinomials etc. but I just can't seem to understand this. If someone could help fill in the missing steps that my solutions manual doesn't include it would be a great help. Even providing an explanation on the steps involved would help me understand the process much better.

    Thanks
     
    Last edited: Aug 26, 2010
  2. jcsd
  3. Aug 26, 2010 #2

    Mark44

    Staff: Mentor

    In the numerator, (x^2 + 9)^2 (-2x) - (9 - x^2)(2)(x^2 + 9)(2x), the two terms have common factors of x^2 + 9 and 2x. If you take these common factors out of the first term, you have 2x(x^2 + 9)[(x^2 + 9)(-1)]. Do the same thing to the second term.
     
  4. Aug 26, 2010 #3
    Thanks for the response: Am I on the right track...

    2x(x^2 + 9)[(x^2 + 9)(-1)] - [(2x)(x^2 + 9) (18 - 2x^2)]

    at this point I'm not clear how the 2x from the second term disappears (i.e is factored out..).
    I understand in what I just did nothing has been factored yet, just rewritten. From this point how does the 2x(x^2 + 9) factor out..

    thanks
     
  5. Aug 26, 2010 #4

    Mark44

    Staff: Mentor

    Now use the idea that a * b + a * c = a(b + c); i.e., the distributive law. Here a = 2x(x^2 + 9)
    2x = 2x * 1, so there's an implied factor of 1 remaining in the second term.
     
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