Solving A Decic Polynomial Equation

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  • čas přidán 5. 07. 2024
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Komentáře • 27

  • @jpolowin0
    @jpolowin0 Před 24 dny +9

    Since all of the terms have even powers, let y = x² to begin with. Then y² + y³ = 2y⁵, or by rearranging and factoring,
    y²(2y³ - y - 1) =0
    This immediately gives y² = x⁴ = 0, x = 0.
    By inspection, y = 1 is a solution, so x² = 1, x = ±1.
    Divide by (y - 1) to get 2y² - 2y - 1 = 0. Solve by quadratic equation to get y = (1 ± 𝒊)/2 = x². Solve that by polar-coordinate stuff which I am too rusty on.

  • @TedHopp
    @TedHopp Před 23 dny +2

    It's worth pointing out that since the original equation is a polynomial of degree 10, there should be 10 solutions. In this problem, you have to count repeated roots, but this lets us know we found them all. (For instance, the common factor of x^4 gives us the solution x=0 with a multiplicity of 4.)

  • @DonRedmond-jk6hj
    @DonRedmond-jk6hj Před 24 dny +1

    Synthetic division gives the cofactors a tad bit faster without having to be clever in your rewriting to use factoring identities.

  • @maxwellarregui814
    @maxwellarregui814 Před 23 dny

    Buenas tares Sres. SyberMath, Gracias es un buen ejercicio. Éxitos.i

  • @scottleung9587
    @scottleung9587 Před 23 dny

    I just substituted y for x^2 and got all the non-trivial solutions from there.

  • @phill3986
    @phill3986 Před 24 dny

    😊😊😊👍👍👍

  • @chasmosaurus3
    @chasmosaurus3 Před 24 dny

    Factor out x^2, it's a little easier than dealing with the quintic.

  • @SidneiMV
    @SidneiMV Před 24 dny +2

    x⁴(2x⁶ - x² - 1) = 0
    x⁴ = 0 => *x = 0*
    2x⁶ - x² - 1 = 0
    x⁶ - x² + x⁶ - 1 = 0
    x²(x⁴ - 1) + (x² - 1)(x⁴ + x² + 1) = 0
    (x² - 1)(x² + 1)x² + (x² - 1)(x⁴ + x² + 1) = 0
    (x² - 1)(x⁴ + x² + x⁴ + x² + 1) = 0
    (x² - 1)(2x⁴ + 2x² + 1) = 0
    x² - 1 = 0 => *x = ± 1*
    2x⁴ + 2x² + 1 = 0
    x² = (-2 ± 2i)/4 = (-1 ± i)/2
    *x = ± √[(-1 ± i)/2]*

  • @DonEnsley-mathdrum
    @DonEnsley-mathdrum Před 24 dny +1

    x ∈ { -1, 0, 1,
    -½ √2√(i-1),
    -½ i √2√(i+1),
    ½ √2√(i-1),
    ½ i √2√(i+1) }
    0 is a quadruple root.(x⁴ = 0)
    10 roots there are.
    x⁴ +x⁶ = 2x¹⁰
    x⁴ [(1+x²)-2x⁶] = 0
    By ZPP x = 0 is a quadruple root.
    1+x²-2x⁶=0
    2x⁶-x²-1=0
    u = x²
    2u³-u-1=0
    u=1 is a root
    x= -1, 1
    2u² +2u+1=0
    ...

  • @charleskrueger5523
    @charleskrueger5523 Před 24 dny

    Your best video so far (and I’ve watched most of them!).

  • @rakenzarnsworld2
    @rakenzarnsworld2 Před 24 dny

    x = 1

  • @lesnyk255
    @lesnyk255 Před 24 dny +1

    "Wait, you forgot..." You'd think by now I'd be hip to this notion of factoring by grouping, but I guess I'm too old school to see it on my own. I'm like that old horse that knows one way back to the barn, and insists on going that way long after the path has been buried under development. At least I've still got enough brain cells left to be delighted watching you do these things. Keep it up, please.

  • @HATTRICK202
    @HATTRICK202 Před 24 dny

    Hello! Could you please come up with a series on Differential Calculus?

  • @1Otnt4354g
    @1Otnt4354g Před 23 dny

    idk man i substitute 1 in and the equation become true. I guess my strat was the fastest then 😅😅😅

  • @braydentaylor4639
    @braydentaylor4639 Před 24 dny

    Isn't it called "dectic", not "decic"?

    • @SyberMath
      @SyberMath  Před 24 dny

      en.wikipedia.org/wiki/Degree_of_a_polynomial

    • @braydentaylor4639
      @braydentaylor4639 Před 24 dny

      @@SyberMath Honestly, "dectic" sounds better

  • @HATTRICK202
    @HATTRICK202 Před 24 dny

    First

  • @ElvisTB
    @ElvisTB Před 24 dny

    Ich habe jetzt mal spontan, bevor ich es mir angesehen habe 1, -1, -i und i. Sollte passen, aber ich lasse mich gern eines besseren belehren.

    • @SidneiMV
      @SidneiMV Před 24 dny

      1 and -1 are right. But i and -i are wrong. And 0 (zero) is also right.

    • @ElvisTB
      @ElvisTB Před 24 dny +1

      @@SidneiMV okay, I see my fault. Didn't consider that ^4 is 1 and ^6 -1.

  • @honestadministrator
    @honestadministrator Před 24 dny

    x^10 - x^4 + x^10 - x^6 = 0
    x^4 [x^6 - 1 + x^6 - x^2] = 0
    x^4 *( x^2 - 1)
    *(x^4 + x^2 + 1 + x^2 * (x^2 + 1)) = 0
    x^4 * ( x^2 - 1) ( x^4 + x^2 + 1/2) = 0
    Hereby
    x = 0, 1, -1