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Math Help - factoring help

  1. #1
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    factoring help

    I am working on being better at factor but I could use some help.

    this (6-3x)^4 + x[4(6-3x)^3(-3)] factors to (6-3x)^3[(6-3x)-12x] to 3(6-3x)^3(2-5x)

    looking for some help on how this is done. thank you.
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  2. #2
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    Re: factoring help

     (6-3x)^4 + x(4(6-3x)^3(-3)) = (6-3x)^3(6-3x-12x) = (6-3x)^3(6-15x) = \cdots

    From here you should notice that you could extract coefficients(common factors) out of the brackets, along with their respective exponents.
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  3. #3
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    Re: factoring help

    I notice a bit of how you are getting that but not entirely. I downloaded an older text book that deals with nothing but factoring. Maybe it'll help but maybe you could explain a little bit. Sorry, I have been stuck teaching everything to myself and sometimes it is a bit difficult. Thank you!
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  4. #4
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    Re: factoring help

    See that  (6-3x) = 3(2-x) \text{ and } (6-15x) = 3 (2-5x) .

    So from where I left off in my last post,

     (6-3x)^3(6-15x)= (3(2-x))^3(3)(2-5x) =3^33^1(2-x)^3(2-5x) = 3^4(2-x)^3(2-5x)=\cdots

    Let me know if you still need some clarification.
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  5. #5
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    Re: factoring help

    hey! I think I got it! thanks so much

    let me go over my steps:

    I start with (6-3x)^4 + x[4(6-3x)^3(-3)]

    take out the binomial (6-3x)^3 giving me (6-3x)^3[(6-3x) + x(4(-3))

    next I have (6-3x)^3[6-3x-12x]

    then, (6-3x)^3[6-15x]

    factor out 3 of the right binomial for 3(6-3x)^3(2-5x)

    then factor out the coefficient of the the left binomial with the exponent to 3^1(3^3)(2-x)(2-5x)

    and a final answer of

    3^4(2-x)(2-5x)
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  6. #6
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    Re: factoring help

    before I get ahead of myself though now I have to factor:

    3(x+4)^2(x-3)^6 + 6(x-3)^5(x+4)^3 down to 3(x+4)^2(x+3)^5(3x+5)
    Last edited by Jonroberts74; September 26th 2013 at 12:36 PM.
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  7. #7
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    Re: factoring help

    nevermind, thank you again!
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