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Math Help - Solve x^4 - x^3 + 8x - 8 = 0, help pls

  1. #1
    zxe
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    Exclamation Solve x^4 - x^3 + 8x - 8 = 0, help pls

    How to solve or factorize this equation?
    x^4 - x^3 + 8x - 8 = 0

    help please
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  2. #2
    Moo
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    Hello,

    You can observe that the sum of the coefficients is 0. So 1 is solution. This means that you can factorise by (x-1).

    So x^4-x^3+8x-8 = (x-1)Q(x)

    Where Q(x) is a polynom of degree 3 -> Q(x)=ax^3 + bx + cx + d, with a,b,c,d to determine by developping (x-1)Q(x).
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  3. #3
    Super Member wingless's Avatar
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    x^4 - x^3 + 8x - 8 = 0

    x^3(x-1) + 8(x-1) = 0

    (x-1)(x^3+8) = 0

    ---> x-1 = 0
    ---> x^3+8 = 0

    x=\{-2,1\}
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wingless View Post
    x^4 - x^3 + 8x - 8 = 0

    x^3(x-1) + 8(x-1) = 0

    (x-1)(x^3+8) = 0

    ---> x-1 = 0
    ---> x^3+8 = 0

    x=\{-2,1\}
    wingless is right. i'd just like to point out that we can solve x^2 + 8 = 0 using the sum of two cubes formula (if you don't see the solution immediately). so, x^3 + 8 = (x + 2)(x^2 - 2x + 2^2) = 0, and now equate each term in the product to zero and solve.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wingless View Post
    x^4 - x^3 + 8x - 8 = 0

    x^3(x-1) + 8(x-1) = 0

    (x-1)(x^3+8) = 0

    ---> x-1 = 0
    ---> x^3+8 = 0

    x=\{-2,1\}
    You missed out on two complex solutions.
    x^4 - x^3 + 8x - 8 = 0

    (x - 1)(x^3 + 8) = 0

    (x - 1)(x + 2)(x^2 - 2x + 4) = 0

    Now let each factor be zero and solve.

    -Dan
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