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
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
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
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.
(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. :-)
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
6a+2b=-11
Lef side is even, right side is odd!
No integer solution!
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
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
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?
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.
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
(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?
ua-cam.com/video/yx2RetjV1Bo/v-deo.html
ua-cam.com/video/55ufNfFofzY/v-deo.html
ua-cam.com/video/B8dCd6PkHMY/v-deo.html
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
(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
bro when are u gonna solve my problem? I need the answer, u explain everything in the best way possible.
Please help me factor 5x^4-7x^3+3x^2-x+1
Hello
X^6............. then (x^6........)(☠☠☠☠☠☠)