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Help solving a third degree polynomial

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data

    x^2(4x + 13) = 9

    2. Relevant equations



    3. The attempt at a solution

    Answers are that x = -3, -1 and 3/4. But how can one compute this other than trial and error?
     
  2. jcsd
  3. Oct 1, 2008 #2

    gabbagabbahey

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    First step is to try expanding it into the form [itex]Ax^3 +Bx^2+Cx+D=0[/itex]. What do you get when you do this?
     
  4. Oct 1, 2008 #3
    4x^3 + 13x^2 +0x -9 = 0
     
  5. Oct 1, 2008 #4

    gabbagabbahey

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    Okay good, you can get rid of the 0x term though :)

    The next step is to recognize that since this is a third degree polynomial, there will be three roots; let's call them [itex]x_1[/itex], [itex]x_2[/itex], and [itex]x_3[/itex]. This means that [itex]4(x-x_1)(x-x_2)(x-x_3)=0=4x^3+13x^2-9[/itex]. Expand the expression on the left. What do you get?
     
  6. Oct 1, 2008 #5

    HallsofIvy

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    If you did not already know the solutions, you could use the "rational root theorem": if x= m/n is a rational number satisfying a polynomial equation then n must evenly divide the leading coefficient and m must evenly divide the constant term.

    Here, those are 4 and -9. The integers that evenly divide 4 are [itex]\pm 1[/itex], [itex]\pm 2[/itex] and [itex]\pm 4[/itex]. The integers that evenly divide 9 are [itex]\pm 1[/itex], [itex]\pm 3[/itex], and [itex]\pm 9[/itex]. That means that the only possible rational roots are [itex]\pm 1[/itex], [itex]\pm 1/2[/itex], [itex]\pm 1/4[/itex], [itex]\pm 3[/itex], [itex]\pm 3/2[/itex], [itex]\pm 3/4[/itex], [itex]\pm 9[/itex], [itex]\pm 9/2[/itex], and [itex]\pm 9/4[/itex]. Check each of those to see if one of them satisfies the equation. (And notice that the solutions you give are among those.)

    See mathworld.wolfram.com/CubicFormula.html

    Of course, there exist cubic equations that have no rational roots. In such a case, you would have to use the (extremely complicated) cubic formula.
     
  7. Oct 1, 2008 #6

    Defennder

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    Hi, gabba just want to make sure I understand your method here. Is it possible that this may not work for some polynomials since we may end up with a cubic equation of a variable say x_1 ? Which of course leads us back to square one.
     
  8. Oct 1, 2008 #7

    gabbagabbahey

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    Yes, you bring up a good point that I had neglected to mention. If the polynomial in question happens to be a perfect cube, then this method won't get you anywhere. However if it is a perfect cube, when you put the polynomial into the form [itex] x^3+C_1x^2+C_2x+C_3[/itex], the constant term [itex]C_3[/itex] will be the cube of the triple root as we can see from the fact that [itex](x-a)^3=x^3-3ax^2+3a^2x-a^3[/itex].
     
  9. Oct 2, 2008 #8

    Defennder

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    But is this guaranteed to work if the polynomial isn't a perfect cube ? ie. One would only have to work with quadratics at most ?
     
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