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3x^2(4x-12)^2 + x^3(2)(4x-12)(4) HALP

  1. Nov 2, 2013 #1
    3x^2(4x-12)^2 + x^3(2)(4x-12)(4)
    Factor this expression completely. This type of question occurs in calculus in using the "product rule".

    Attempt:
    I first factor out a x^2 (4x-12) factor out the 4 from this, then have 4x^2 (X-3)
    The left overs are: 3(4x-12) + x(2)(4)
    I factor out a 4 from the left overs, I now have. 16x^2(x-3) 3(x-3) + x(2)
    Then I simplify: 16x^2 (x-3) 3(x-3) +2x = 16x^2 (x-3) (5x-9)


    Which is the right answer, but I may have been biased into finding this as I did know the answer before hand. So is my reasoning sound? Maybe you could go through this yourself and show me your process, practice does not hurt :)
     
  2. jcsd
  3. Nov 2, 2013 #2

    UltrafastPED

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    When you factor out x^2 (4x-12) you have:
    [x^2 (4x-12)]*[3(4x-12) + 8x] ... which is what you got. Then you factor out the 4:
    [16 x^2 (x-13)]*[3(x-3)+2x] => [16 x^2 (x-3)]*[3x - 9 +2x] => [16 x^2 (x-3)]*[5x-9]
    And then discard the extra brackets.

    Like you, I did not notice the extra factor of 4 in (4x-12) on the first pass.
     
  4. Nov 2, 2013 #3
    Thanks for the reply. I have another question...

    Factor this completely.

    4a^2c^2 - (a^2 - b^2 + c^2)^2

    I am so stumped on this one.

    Thanks again for the previous answer.
     
  5. Nov 2, 2013 #4

    UltrafastPED

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    Square the second term ... you should get six terms. One of these six can be combined with the first term of the original expression.

    Then look for simplifications ...
     
  6. Nov 3, 2013 #5

    epenguin

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    You often get things like this in exercises and even sometimes in real science problems. This is a "difference of two squares". You've probably seen the factorising of that. After using that there might be some further simplification possible.
     
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