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Finding solutions to a cubic

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the real solutions of [itex]x^3+6x^2-8x+1[/itex]

    2. Relevant equations
    quadratic


    3. The attempt at a solution
    First I tried simple numbers starting with 1 and found that +1 was a solution and therefore I could write

    [tex]
    x^3+6x^2-8x+1=(x-1)(Ax^2+Bx+C) \\
    x^3+6x^2-8x+1=Ax^3+Bx^2+Cx-Ax^2-Bx-C \\
    [/tex]

    And then equated the coeffeciants

    For x^3, [itex]1=A[/itex]
    For x^2, [itex]6=B-A[/itex]
    For x^1, [itex]-8=C-B[/itex]
    For x^0, [itex]1=-C[/itex]

    With knowing that A=1 and C= -1 , B can be found to be 7 .

    And therefore I can write [itex]x^3+6x^2-8x+1=(x-1)(x^2+7x-1)[/itex]

    If I then use the quadratic formula on the quadratic factor I end up with
    [itex]\frac{-7\pm\sqrt{53}}{2}[/itex]

    And as 53 is prime that is as far as it can go without losing accuracy.

    I dont know if this is all I have to do, if I have done what I have done correctly or what. I am a bit confused. Any help/advice appreciated.
     
  2. jcsd
  3. Oct 6, 2013 #2

    Office_Shredder

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    That looks entirely correct, and is all you have to do. You can include a line at the end that says

    Therefore, the solutions are [itex] x=1,\ x=\frac{-7+\sqrt{53}}{2}[/itex] and [itex] x=\frac{-7-\sqrt{53}}{2}[/itex]
     
  4. Oct 6, 2013 #3

    LCKurtz

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    I'm curious why you didn't just divide the cubic by ##x-1## using either long division or, preferably, synthetic division to get the quadratic factor. Much less work...
     
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