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Math Help - Factoring Quartic equation

  1. #1
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    Factoring Quartic equation

    I really don't understand how they factorise this:

    y^4+y^2+1

    Can anyone lend me a hand?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nabbiechan View Post
    I really don't understand how they solve this:

    y^4+y^2+1

    Can anyone lend me a hand?
    Can you solve y^2=z. Then, this becomes z^2+z+1=0. Can you solve this?
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  3. #3
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    Oh sorry, I put solve instead of factorise. I apologize! >_<
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nabbiechan View Post
    Oh sorry, I put solve instead of factorise. I apologize! >_<
    Factor over the reals? Solve z^2+z+1=0 and then use this to solve y^4+y^2+1=0 and then use this to factor it into linear factors.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    z^2+z+1=0 and then use this to solve y^4+y^2+1=0
    This is the part I'm really confused about.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nabbiechan View Post
    This is the part I'm really confused about.
    Which part?
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  7. #7
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    y^4 + y^2 + 1.

    Make the substitution x = y^2 so that the expression becomes

    x^2 + x + 1

    = x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1

     = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.


    This is not factorisable over the Reals.
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  8. #8
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    Quote Originally Posted by Drexel28 View Post
    Which part?
    where you're going to use z^2+z+1 to solve the equation.

    @prove it:

    I only was able to solve it through the calculator and it's factorisable, but I wanted to know the method. The answer was [LaTeX ERROR: Convert failed] . A friend told me to use another equation similar to it which made it more confusing.
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  9. #9
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    Quote Originally Posted by nabbiechan View Post
    I really don't understand how they factorise this:

    y^4+y^2+1

    Can anyone lend me a hand?
    hi

    y^4+y^2+1

    =y^4+(2y^2-y^2)+1

    =(y^4+2y^2+1)-y^2

    =(y^2+1)^2-y^2

    =(y^2+1-y)(y^2+1+y)

    using the fact that a^2-b^2=(a+b)(a-b)

    ok ?
    Last edited by mathaddict; April 7th 2010 at 12:09 AM.
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  10. #10
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    Quote Originally Posted by mathaddict View Post
    hi

    y^4+y^2+1

    =y^4+(2y^2-y^2)+1

    =(y^4+2y^2+1)-y^2

    =(y^2+1)-y^2

    =(y^2+1-y)(y^2+1+y)

    using the fact that a^2-b^2=(a+b)(a-b)

    ok ?

    ooh. okay, thanks for everyone's help!
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