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Math Help - Algebra 2 help..

  1. #1
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    Algebra 2 help..

    3x^3 -4x^2-17x+6=0

    I'm just told to solve it, it's in a chapter about polynomials. I know the answers, 3,1/3, and -2. I just don't know how to get them.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by MadisonDakota View Post
    3x^3 -4x^2-17x+6=0

    I'm just told to solve it, it's in a chapter about polynomials. I know the answers, 3,1/3, and -2. I just don't know how to get them.
    there are ways how to solve it..

    try using rational root test and synthetic division..
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  3. #3
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    Another hint.

    When x=-2, 3(-2)^3-4(-2)^2-17(-2)+6=0

    therefore, 3x^3-4x^2-17x+6 = (x+2)(ax^2+bx+c)

    Hope it help you to solve the problem now
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  4. #4
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    But I can't use -2 if I didn't get the answer from the teacher. Like how do I do it, w/o using one of the answers?
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  5. #5
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    By trial and error, you have to determine one of the roots first. then try to solve the problem. Let me give you the answer you try to work it out.

    (x+2)(3x^2-10x+3)
    (x+2)(3x-1)(x-3)

    Hope it helps.
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by tester85 View Post
    By trial and error, you have to determine one of the roots first. then try to solve the problem. Let me give you the answer you try to work it out.

    (x+2)(3x^2-10x+3)
    (x+2)(3x-1)(x-3)

    Hope it helps.
    trial and error doesn't make sense when you are talking on infinitely many numbers to try on.. i mean, it is a waste of time.. try decreasing the number of numbers you are going to try,,

    the rational root test tells you that if you have a polynomial of degree n whose leading coefficient is a_n and whose constant term is a_0, then the possible roots of the polynomial is given by \frac{p}{q} where p is a factor of a_0 and q is a factor of a_n

    in your example, a_0 = 6 and a_n = 3

    so p can be equal to \pm 1, \pm 2, \pm 3, \pm 6
    and q can be equal to \pm 1, \pm 3

    from here, you may start your trial and error by choosing any combination of \frac{p}{q}

    after getting one, you may do previous poster's suggestion by taking out and factoring the quadratic factor..
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