How to factor a 5-term polynomial (the double-cross method)
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- Опубликовано: 2 июн 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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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
6a+2b=-11
Lef side is even, right side is odd!
No integer solution!
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
could you do the Δ(4th/D) problem from the Greek panhellenics 2024 if possible?
(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. :-)
Do you have a video on lots of different ways to factor different polynomials?
ruclips.net/video/yx2RetjV1Bo/видео.html
ruclips.net/video/55ufNfFofzY/видео.html
ruclips.net/video/B8dCd6PkHMY/видео.html
(x−1)(x−3)(x2−2x+4) by long division
true, but to do long division don't you need to first guess either (x-1) or (x-3)?
@@z000ey this how I got (x-1) and (x-3) then I did long division and got my answer.
But you can't always do that
2:30 why do a and b need to be whole numbers? Is there a rule for that?
Being a monic equation with integer coefficients, it can't have factors with fraccional coeficients
this is harder than the synthetic division method, i don't this i can use this method efficiently, i am too dumb
me too
Isnt synthetic division only for cubics?
@@Orillians It's actually for any polynomial
@@bruhbro9813 damn thank you!
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.
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
bro when are u gonna solve my problem? I need the answer, u explain everything in the best way possible.
Hello
X^6............. then (x^6........)(☠☠☠☠☠☠)