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Homework Help: Complex roots

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    6x^3 - 3x^2 - 45x

    2. Relevant equations

    3. The attempt at a solution
    -first factor out 3x
    3x(2x^2 - x -15)

    2 x 5 = 30 so...

    3x(2x^2 + 5x - 6x - 15)

    -separate the equations
    3x(2x^2 + 5x)( - 6x -15)

    -simplify a few things.
    3x^2(2x + 5) -3 (2x + 5)

    (3x^2 - 3)(2x + 5)

    set both equations to zero
    2x + 5 = 0
    2x = -5
    x = -5/2

    3x^2 - 3 = 0
    3x^2 = 3

    thats where I get stuck, that is if I did this right. thanks for any help you can offer.
  2. jcsd
  3. Oct 22, 2007 #2


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    Staff Emeritus
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    What've you done to get from the line above to this:
    It's incorrect, anyway. You should try using the quadratic formula to factorise (2x^2 -x- 15)
  4. Oct 23, 2007 #3


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    Wasn't this supposed to be an equation? And, if so, doesn't it need an "=" somewhere? Did you mean 6x^3- 3x^2- 45x= 0?

    Well, no, 2 x 5= 10, not 30. But there is no 30 in the equation anyway so I don't know what you were trying to do!

    ?? Did you mean 3x[(2x^2+ 5x)+ (-6x-15)]?

    Again, you mean 3x^2[(2x+5)- 3(2x+5)]

    No, since the 3x is multiplied by both 2x+ 5 and -3(2x+5) you cannot separate the expression like that.
    Well, obviously that gives x^2= 1 which has solutions x= 1 and x= -1- but they obviously do not satisfy the original equation. Go back to 3x^2(2x^2 - x -15)= 0. That gives you one obvious solution. Now try to factor 2x^2- x- 15.

    However, this equation does not has complex roots!
    Last edited by a moderator: Oct 23, 2007
  5. Oct 24, 2007 #4
    ok thanks a lot. that really does help a lot. My tutor also pointed out some of my mistakes, but with the combination of that and this I'm very clear on the subject. thanks again
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