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Math Help - Help solve quartic

  1. #1
    Senior Member DivideBy0's Avatar
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    Help solve quartic

    x^4 - 4x^3 + 7x^2 + 22x + 24 = 0

    I have no idea how to solve this, it suddenly popped up in the quadratic section of the book. Help please!
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  2. #2
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    Edit nevermind, there doesn't seem to be an exact value from polynomials O_O
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    x^4 - 4x^3 + 7x^2 + 22x + 24 = 0

    I have no idea how to solve this, it suddenly popped up in the quadratic section of the book. Help please!
    r_maths is right i think, there are no nice numbers for this polynomial, i don't even think it has roots. are you sure you typed it correctly?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    x^4 - 4x^3 + 7x^2 + 22x + 24 = 0

    I have no idea how to solve this, it suddenly popped up in the quadratic section of the book. Help please!
    We can easily show that all the zeros of this polynomial are complex, since the graph of it never crosses the x-axis.

    The best we can do (if we don't use the general formula for a quartic, which is "butt ugly") is to try to find quadratic factors of the form:
    (x - (a + Ib))(x - (a - Ib))

    I messed with this for a while, but couldn't come up with anything intelligible.

    -Dan
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    Forum Admin topsquark's Avatar
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    I have been able to show the polynomial does not have factors of the form
    (x^2 + ax + b)(x^2 + cx + d)
    where a, b, c, d are all integers.

    With that said, I'm giving up!

    -Dan
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  6. #6
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    Re:

    Re:

    It doesn't appear that this equation intersects the x-axis at all...
    Attached Thumbnails Attached Thumbnails Help solve quartic-graph-4.jpg  
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  7. #7
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    I do not think there is nice way to solve this without the quartic formula. If instead of the 7 you have a -7 it would work.

    Let me explain. Some of these polynomials can be solved by manipulating them moving factors until you bring it to the form you can solve it in.

    I was able to show this polynomial is irreducible over Q. Meaning we cannot achieve this with doing nice moves, i.e. moving factors around if you know what I mean.
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