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Math Help - Solving a polynomial equation

  1. #1
    tac
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    Solving a polynomial equation

    I feel like this should be so easy, but I am stumped. I have to solve for y^3 = y.

    This is how I am doing, but do not think it is correct:

    y^3 = y
    -y -y

    y^3 - y = 0

    y (y^2 - y) = 0

    y^2 - y = 0
    +y +y

    y^2 = y
    y y

    y = 1

    Solution:
    y = 0
    y = 1

    Any help would be appreciated. Thank you!
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  2. #2
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    Re: Solving a polynomial equation

    y^3 - y = 0

    y(y^2-1) = 0

    y(y-1)(y+1) = 0
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  3. #3
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    Re: Solving a polynomial equation

    Quote Originally Posted by tac View Post
    ...
    y^3 - y = 0

    y (y^2 - y) = 0
    ...
    That was a mistake in factoring, which is why you "lost" the solution y=-1. Skeeter's post shows the correct factorization.

    Also, when you solve for zeros of a polynomial, it's best to do the entire factorization (with 0 on the right hand side) first, and then read off the roots (or use quadratic equations to get the roots of a quadratic that didn't factor nicely).
    Picking off roots by dividing and manipulating as you go - the way you were working the problem - is not a good approach for many reasons. Again, see Skeeter's post for the correct way to do this.
    Last edited by johnsomeone; October 7th 2012 at 08:06 AM.
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  4. #4
    tac
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    Re: Solving a polynomial equation

    Thank you!

    Those perfect squares get me every time.

    After factoring out the perfect square I am getting:

    y = 0
    y = 1
    y = -1

    Can there be three answers? When I plug in each one they all work.
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  5. #5
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    Re: Solving a polynomial equation

    Yes! Degree 3 means 3 solutions are possible.
    There could be less than 3 answers, but NEVER more (well, except for the degenerate polynomial 0).
    Degree n means n solutions are possible, but never more. There are never n+1 different roots for an n-th degree polynomial. This is counting complex roots.

    Ex:
    y^2-3y+2 = 0 has two solutions (y = 1, y=2).
    y^2+1 = 0 has no real solutions, but two complex solutions (i and -i).
    y^2 = 0 has only one solution (real or complex), which is y = 0.
    Likewise
    y^2 -2y + 1 = 0 has only one solutions (real or complex), which is y = 1.
    Last edited by johnsomeone; October 7th 2012 at 08:18 AM.
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