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Math Help - hard quartic equation problem.

  1. #1
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    hard quartic equation problem.

    this one is giving me some trouble. I am not sure what method to use.



    I figure first expand the equation which gives me quartic form.



    but after that I am stuck. I don't think I can factor with the methods I know. I tried a few ways. and substitution seems to fail to get anywhere.

    the solution is which does expand to the the same quartic as above. I have NO idea how to do this.
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  2. #2
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    Let .

    Substituting into your Quartic gives

    .

    Expanding this out and collecting like terms will give you the required form.
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    I tried substitution but not the way you did it in the quartic thanks!

    how did you know to use (u+1)?
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    You let u equal whatever you want your function to be a function of, in this case, x - 1.
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    I see, thanks.
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  6. #6
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    Hello, skoker!


    We have: .(3x + 2)
    [(x-1)^3 + 3(x-1)^2 + 2(x-1) + 1]

    . . . . . = .
    [3(x-1) + 5] [(x-1)^3 + 3(x-1)^2 + 2(x-1) + 1]

    . . . . . = . 3(x-1)^4 + 9(x-1)^3 + 6(x-1)^2 + 3(x-1)
    . . . . . . . . . . . . . . + 5(x-1)^3 + 15(x-1)^2 + 10(x-1) + 5

    . . . . . = . 3(x-1)^4 + 14(x-1)^3 + 21(x-1)^2 + 13(x-1) + 5

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  7. #7
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    thank you soroban. that is close to how I was trying to do substitution, before distribution. but how do you get the 5 in [3(x-1)+5]?
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