Solving A Decic Polynomial Equation

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  • Опубликовано: 5 июл 2024
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Комментарии • 27

  • @jpolowin0
    @jpolowin0 27 дней назад +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 27 дней назад +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 27 дней назад +1

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

  • @maxwellarregui814
    @maxwellarregui814 26 дней назад

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

  • @scottleung9587
    @scottleung9587 27 дней назад

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

  • @chasmosaurus3
    @chasmosaurus3 27 дней назад

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

  • @phill3986
    @phill3986 28 дней назад

    😊😊😊👍👍👍

  • @rakenzarnsworld2
    @rakenzarnsworld2 27 дней назад

    x = 1

  • @SidneiMV
    @SidneiMV 28 дней назад +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 27 дней назад +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 28 дней назад

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

  • @lesnyk255
    @lesnyk255 28 дней назад +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.

  • @1Otnt4354g
    @1Otnt4354g 27 дней назад

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

  • @HATTRICK202
    @HATTRICK202 28 дней назад

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

  • @braydentaylor4639
    @braydentaylor4639 27 дней назад

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

    • @SyberMath
      @SyberMath  27 дней назад

      en.wikipedia.org/wiki/Degree_of_a_polynomial

    • @braydentaylor4639
      @braydentaylor4639 27 дней назад

      @@SyberMath Honestly, "dectic" sounds better

  • @HATTRICK202
    @HATTRICK202 28 дней назад

    First

  • @ElvisTB
    @ElvisTB 28 дней назад

    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 28 дней назад

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

    • @ElvisTB
      @ElvisTB 28 дней назад +1

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

  • @honestadministrator
    @honestadministrator 28 дней назад

    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