How to factor a 5-term polynomial (the double-cross method)

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  • Опубликовано: 13 дек 2024

Комментарии • 29

  • @nevoitzhak2092
    @nevoitzhak2092 6 месяцев назад +7

    The sum of the coefficients is 0 therefore 1 is a solution
    X⁴-4x³+2x²-11x+12=
    X⁴-x³-3x³+3x²-x²+x-12x+12=
    X³(x-1)-3x²(x-1)-x(x-1)-12(x-1)=
    (X-1)(x³-3x²-x-12)
    Then factoring a cubic is much easier

  • @Iomhar
    @Iomhar 9 месяцев назад +10

    6a+2b=-11
    Lef side is even, right side is odd!
    No integer solution!

  • @JoseAntonio-ng5yu
    @JoseAntonio-ng5yu 6 месяцев назад

    If the independent term has too many factors, this can be quite hard, as you must check every combination of two factors 2 times each, as they can also have switched signs.
    The solution of the system of equations is a=(Ac-C)/(c-d) and b=(C-Ad)/(c-d). A and C are the coefficients of x^3 and x, and c and d are the top and bottom numbers at the right cross which are factors of the independent term. Substitute directly into a to eliminate more quickly those in which a is not a in integer. If a is an integer, obtain b and check the last condition

  • @major__kong
    @major__kong 6 месяцев назад +1

    (x - r1)(x - r2)(x - r3)(x - r4) = ax^4 + bx^3 + cx^2 + dx + e
    Multiply out the left side, set the coefficients of like powers of x equal to each other, solve the system of equations for a, b, c, d, and e in terms of r1, r2, r3, and r4. Hint: a = 1. :-)

  • @a_man80
    @a_man80 3 месяца назад +2

    My solution: x^3*(x-4)+(2x-3)(x-4)=(x-4)(x^3+2x-3)=(x-4)(x-1)(x^2+x+3) first factor out x^3 from first two terms, then factorize remaining second degree polynomial, then factor out (x-4), i found the x-1 factor by trying in my head

  • @giorgouis9642
    @giorgouis9642 6 месяцев назад +1

    could you do the Δ(4th/D) problem from the Greek panhellenics 2024 if possible?

  • @lushleafy1174
    @lushleafy1174 9 месяцев назад +6

    this is harder than the synthetic division method, i don't this i can use this method efficiently, i am too dumb

    • @apotatoman4862
      @apotatoman4862 9 месяцев назад

      me too

    • @Orillians
      @Orillians 6 месяцев назад

      Isnt synthetic division only for cubics?

    • @bruhbro9813
      @bruhbro9813 6 месяцев назад +1

      ​@@Orillians It's actually for any polynomial

    • @Orillians
      @Orillians 6 месяцев назад

      @@bruhbro9813 damn thank you!

    • @goliath6278
      @goliath6278 6 месяцев назад +3

      Not all 4th degree polynomials can be factored with synthetic division. If all the roots are imaginary, then synthetic division can't help, and a method like this might be the only way.

  • @dutchie265
    @dutchie265 6 месяцев назад

    2:30 why do a and b need to be whole numbers? Is there a rule for that?

    • @JoseAntonio-ng5yu
      @JoseAntonio-ng5yu 6 месяцев назад +3

      Being a monic equation with integer coefficients, it can't have factors with fraccional coeficients

  • @Ben_Long
    @Ben_Long 6 месяцев назад

    Do you have a video on lots of different ways to factor different polynomials?

    • @NadiehFan
      @NadiehFan 6 месяцев назад

      ruclips.net/video/yx2RetjV1Bo/видео.html
      ruclips.net/video/55ufNfFofzY/видео.html
      ruclips.net/video/B8dCd6PkHMY/видео.html

  • @wdobni
    @wdobni 6 месяцев назад

    you should be able to calculate whether useful quantum computers will ever be developed....the limiting factor is noise and we know how much noise there is and how much each unit of noise reduces quantum integrity with respect to accuracy.....it should be a straightforward matter to compute whether noise will ever be sufficiently mastered to permit reliable useful quantum computing beyond the toy stage

  • @HoussamMoghrabi
    @HoussamMoghrabi 6 месяцев назад

    (x−1)(x−3)(x2−2x+4) by long division

    • @z000ey
      @z000ey 6 месяцев назад

      true, but to do long division don't you need to first guess either (x-1) or (x-3)?

    • @HoussamMoghrabi
      @HoussamMoghrabi 6 месяцев назад

      @@z000ey this how I got (x-1) and (x-3) then I did long division and got my answer.

    • @JoseAntonio-ng5yu
      @JoseAntonio-ng5yu 6 месяцев назад

      But you can't always do that

  • @kadinatorgaming
    @kadinatorgaming 3 месяца назад

    Please help me factor 5x^4-7x^3+3x^2-x+1

  • @aneeshbro
    @aneeshbro 6 месяцев назад +1

    bro when are u gonna solve my problem? I need the answer, u explain everything in the best way possible.

  • @nuctang
    @nuctang 6 месяцев назад

    X^6............. then (x^6........)(☠☠☠☠☠☠)

  • @inyomansetiasa
    @inyomansetiasa 6 месяцев назад

    Hello