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Math Help - Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

  1. #1
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    Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    So I am working my way through an algebra practice book, and I am stumped on this question:

    4x^4 + (3x^2)(y^2) + y^4

    Now the answer is listed as (2x^2 + xy + y^2)(2x^2 - xy + y^2)

    What steps does one take to obtain this answer?

    Thanks
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  2. #2
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    Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    We have two fourth powers and one sqare beneath monomyals Let us try enshorted multiplyig formulas
    (2x^{2}+x^{2})^{2}=4x^{4}+4x^{2}y^{2}+y^{4}=4x^{4}  +3x^{2}y^{2}+x^{2}y^{2}
    if we odd bothsidely x^{2}y^{2} we have
    (2x^{2}+x^{2})^{2}-x^{2}y^{2}=our formula
    Last edited by topsquark; August 22nd 2014 at 06:32 AM.
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  3. #3
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    Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    Quote Originally Posted by Cartesius24 View Post
    We have two fourth powers and one sqare beneath monomyals Let us try enshorted multiplyig formulas
    (2x^{2}+x^{2})^{2}=4x^{4}+4x^{2}y^{2}+y^{4}=4x^{4}  +3x^{2}y^{2}+x^{2}y^{2}
    if we odd bothsidely x^{2}y^{2} we have
    (2x^{2}+x^{2})^{2}-x^{2}y^{2}=our formula
    Thanks, but I am still confused. What steps do you use to derive that answer? Also, you wrote initially (2x^2 + x^2)^2, did you mean (2x^2 + y^2)^2?

    Another example is the following problem:

    x^8 - 12x^4 + 16

    Do I just add and subtract x^4 from this? I find that the two factors of 16 are -4 and -4, so if I add and subtract 4x^4 from the problem, I get (x^8 - 8x^2 + 16) - 4x^2. This I can then factor into the answer.

    I used some guesswork to get this though, by knowing that (x^4 - 4)^2 is x^8 - 8x^4 + 16. So I assumed then that since -12x^4 + 4x^4 is -8x^4, then I then assumed I subtract 4x^4 from the other side.

    Are problems like this solved by some guessing/trial-and-error, or is there a specific step I am missing?
    Last edited by EngineMan; August 22nd 2014 at 12:59 PM.
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  4. #4
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    Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    We can also factor like this

    x^8-12x^4+16=\left(x^2+1-\sqrt{5}\right) \left(x^2-1-\sqrt{5}\right) \left(x^2-1+\sqrt{5}\right) \left(x^2+1+\sqrt{5}\right)

    or like this

    x^8-12x^4+16=\left(x^4-6-\sqrt{20}\right)\left(x^4-6+\sqrt{20}\right)

    or as a product of linear factors
    Last edited by Idea; August 22nd 2014 at 09:22 PM.
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    Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    Quote Originally Posted by Idea View Post
    We can also factor like this

    x^8-12x^4+16=\left(x^2+1-\sqrt{5}\right) \left(x^2-1-\sqrt{5}\right) \left(x^2-1+\sqrt{5}\right) \left(x^2+1+\sqrt{5}\right)

    or like this

    x^8-12x^4+16=\left(x^4-6-\sqrt{20}\right)\left(x^4-6+\sqrt{20}\right)

    or as a product of linear factors
    Where do the \sqrt{5} and the \sqrt{20} come from?
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  6. #6
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    Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    First, factor the quadratic polynomial

    u^2-12u+16=\left(u-6+\sqrt{20}\right)\left(u-6-\sqrt{20}\right)

    Then let u=x^4 to get

    x^8-12x^4+16=\left(x^4-6+\sqrt{20}\right)\left(x^4-6-\sqrt{20}\right)

    Next,

    \sqrt{6+\sqrt{20}}=1+\sqrt{5}
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    Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

    Quote Originally Posted by Idea View Post
    First, factor the quadratic polynomial

    u^2-12u+16=\left(u-6+\sqrt{20}\right)\left(u-6-\sqrt{20}\right)

    Then let u=x^4 to get

    x^8-12x^4+16=\left(x^4-6+\sqrt{20}\right)\left(x^4-6-\sqrt{20}\right)

    Next,

    \sqrt{6+\sqrt{20}}=1+\sqrt{5}
    I think this might be a bit beyond where I am at right now, as I am not yet at quadratic equations in my studying.
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