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

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  • เผยแพร่เมื่อ 15 ต.ค. 2024
  • Algebra tutorial on factoring a 5-term polynomial x^4-4x^3+2x-11x+12 using the double-cross method. Factoring polynomials is crucial for solving polynomial equations for your algebra and precalculus classes. Check out a harder double-cross factoring problem: • How to factor a hard 4...
    Check out different methods of factoring this 5-term polynomial below.
    By rational zero theorem: • How to factor a 5-term...
    By grouping: • How to factor a 5-term...
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ความคิดเห็น • 27

  • @Iomhar
    @Iomhar 7 หลายเดือนก่อน +10

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

  • @nevoitzhak2092
    @nevoitzhak2092 4 หลายเดือนก่อน +6

    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

  • @JoseAntonio-ng5yu
    @JoseAntonio-ng5yu 4 หลายเดือนก่อน

    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

  • @a_man80
    @a_man80 หลายเดือนก่อน +1

    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 4 หลายเดือนก่อน +1

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

  • @major__kong
    @major__kong 4 หลายเดือนก่อน +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. :-)

  • @lushleafy1174
    @lushleafy1174 7 หลายเดือนก่อน +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 7 หลายเดือนก่อน

      me too

    • @Orillians
      @Orillians 4 หลายเดือนก่อน

      Isnt synthetic division only for cubics?

    • @bruhbro9813
      @bruhbro9813 4 หลายเดือนก่อน +1

      ​@@Orillians It's actually for any polynomial

    • @Orillians
      @Orillians 4 หลายเดือนก่อน

      @@bruhbro9813 damn thank you!

    • @goliath6278
      @goliath6278 4 หลายเดือนก่อน +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.

  • @Ben_Long
    @Ben_Long 4 หลายเดือนก่อน

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

    • @NadiehFan
      @NadiehFan 4 หลายเดือนก่อน

      th-cam.com/video/yx2RetjV1Bo/w-d-xo.html
      th-cam.com/video/55ufNfFofzY/w-d-xo.html
      th-cam.com/video/B8dCd6PkHMY/w-d-xo.html

  • @HoussamMoghrabi
    @HoussamMoghrabi 4 หลายเดือนก่อน +1

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

    • @z000ey
      @z000ey 4 หลายเดือนก่อน

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

    • @HoussamMoghrabi
      @HoussamMoghrabi 4 หลายเดือนก่อน

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

    • @JoseAntonio-ng5yu
      @JoseAntonio-ng5yu 4 หลายเดือนก่อน

      But you can't always do that

  • @kadinatorgaming
    @kadinatorgaming หลายเดือนก่อน

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

  • @dutchie265
    @dutchie265 4 หลายเดือนก่อน

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

    • @JoseAntonio-ng5yu
      @JoseAntonio-ng5yu 4 หลายเดือนก่อน +3

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

  • @wdobni
    @wdobni 4 หลายเดือนก่อน

    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

  • @aneeshbro
    @aneeshbro 4 หลายเดือนก่อน +1

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

  • @inyomansetiasa
    @inyomansetiasa 4 หลายเดือนก่อน

    Hello

  • @nuctang
    @nuctang 4 หลายเดือนก่อน

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