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Constructible root from cubic polynomial

  1. Apr 29, 2009 #1
    Does x^3 - 3x + 3sqrt(3) = have a constructible root?

    my solution:
    suppose a is a constructible root of the equation above.
    we square both sides to get x^6 - 6x^4 + 9x^2 = 27.
    since a is constructible, a^2 is constructible as well and we can turn this equation into cubic poly with rational coefficients, and it becomes y^3 - 6y^2 + 9y - 27 = 0.

    If this cubic poly has a constructible root, it must have a rational root in form of m/n.
    (m/n)^3 - 6(m/n)^2 + 9(m/n) = 27.

    How do I proceed from here? Detailed steps would be appreciated..
     
  2. jcsd
  3. May 1, 2009 #2
    The form will wind up as m^3-6m^2n+9mn^2-27 n^3 =0. This is supposed to be a solution. (We assume that m and n are relatively prime, that is, contain no common factors.) So look at m, there is only 1 term in which it is not present.

    So take it from there!
     
  4. May 1, 2009 #3

    HallsofIvy

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    If it has a rational root, m/n, then m must divide the constant term and n must divide the leading coefficient. Here the constant term is -27 so m must be 1, -1, 3, -3, 9, -9, 27, or -27. The leading coefficient is 1 so the n must be either 1 or -1 which means the only possible rational roots are 1, -1, 3, -3, 9, -9, 27, or -27. Do any of those satisfy the equation?
     
  5. May 2, 2009 #4
    You can apply the Rational Roots Theorem to

    y^3 - 6y^2 + 9y - 27 = 0.

    as HallsofIvy explained, but it is possible to reduce the number of root candidates by substitution.

    Define

    P(y) = y^3 - 6y^2 + 9y - 27

    P(1) = -23, so y = 1 is not a root. Then, instead of trying out the other candidates, you can look at the polynomial:

    Q(t) = P(1+t)

    The coefficient of t^3 in Q(t) is 1 and the constant term in Q(t) is
    Q(0) = P(1) = -23

    This means that any rational root of Q(t) must be a divisor of 23, so the possible roots are:

    t = 1, -1, 23, -23

    The possible rational roots of P(y) are then the values of 1+t which are:

    y = 2, 0, 24, -21

    But any rational root must also be in the list give by HallsofIvy. Since none of the values for y listed above are on that list, there are no rational roots.
     
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