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Math Help - Simple Rearanging to solve for x

  1. #1
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    Simple Rearanging to solve for x

    Okay so I have to solve for x


    120 = 2(x-2)(x-3)(x+5)<br />
    60 = (x-2)(x-3)(x+5)
    60 = x^3 - 19x + 30

    I am stuck from here. I don't know if I should have expanded or not.
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  2. #2
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    Quote Originally Posted by JimmyRP View Post
    Okay so I have to solve for x


    120 = 2(x-2)(x-3)(x+5)<br />
    60 = (x-2)(x-3)(x+5)
    60 = x^3 - 19x + 30

    I am stuck from here. I don't know if I should have expanded or not.
    x^3 - 19x - 30 = 0

    rational root theorem ... try x = -2 as your first root, do some synthetic division, and solve for the other two roots.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by JimmyRP View Post
    Okay so I have to solve for x


    120 = 2(x-2)(x-3)(x+5)<br />
    60 = (x-2)(x-3)(x+5)
    60 = x^3 - 19x + 30

    I am stuck from here. I don't know if I should have expanded or not.

    now subtract 60 from both sides

    x^3-19x-30=0

    Now we need to use the rational roots theorem. So we know the only roots are factors of 30.

    So if we try x=5 we get

    5^3-19(5)-30=125-95-30=0

    so 5 is a root so x-5 is a factor. So by long division we get

    (x-5)(x^2+5x+6)=0

    (x-5)(x+2)(x+3)=0

    I hope this helps
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  4. #4
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    thanks a lot. I just realized what I had to do, then I checked here and you guys helped. Thanks a lot
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