MHB Unraveling the Complexities of Factorization in Multivariable Polynomials

  • Thread starter Thread starter bergausstein
  • Start date Start date
  • Tags Tags
    Factorization
AI Thread Summary
The discussion revolves around the factorization of a complex multivariable polynomial, initially presented with a typo that complicated the process. After correcting the expression, a user found the factored form using Wolfram Alpha, which is expressed as two polynomial factors. Participants expressed frustration over the difficulty of manual factorization and sought clarification on how to approach such problems without computational tools. The conversation highlighted the challenges of polynomial factorization, emphasizing that advanced techniques and trial-and-error methods are often required. Ultimately, the discussion underscores the complexity of factorization in multivariable polynomials and the need for deeper mathematical understanding.
bergausstein
Messages
191
Reaction score
0
factor

$\displaystyle 24x^4y^2+28xy^3+30x^5y^2-72xy-6x^5+35x^2y^3-18x^6y-32x^3y+33x^3y^2+63y^3+10x^4y+24x^4-14x^2y$

i have no idea where to start please help me.
 
Last edited:
Mathematics news on Phys.org
no one will help you. because it's tedious and impossible. :p
 
Are you sure you've copied the expression correctly?
 
uhm oh. there's a typo. $24x^4y^2$ should be $24x^4$ sorry. i edited now.
 
bergausstein said:
uhm oh. there's a typo. $24x^4y^2$ should be $24x^4$ sorry. i edited now.

W|A still finds no factored form.
 
I'm really sorry. i guess i edited it correctly this time. please bear with me.
 
bergausstein said:
I'm really sorry. i guess i edited it correctly this time. please bear with me.

Okay now it factors. This is what W|A returns:

$$-\left(3x^3-5x^2y-4xy-9y \right)\left(6x^3y+2x^2-8x+7y^2 \right)$$

I suggest looking at expanding that to gain insight into how it factors.
 
yes this the answer. but I'm having a hard time understanding how did you arrive at the answer. please help. anybody?
 
Last edited:
I don't see any obvious factorization of your polynomial, neither any homogeneous-inhomogeneous trick for this one (This is provable). There are not much of polynomial-factorization algorithm and the best is not something pre-algebra level student can understand. From where did you found this one out?
 
  • #10
this is hard.
 
Last edited:
  • #11
our instructor gave us this problem. i don't know from where did he get this.

i want to learn how to solve this. I've used all the method i know to solve this but no success.

please help me.
 
Last edited:
  • #12
We are given to factor:

$$24x^4y^2+28xy^3+30x^5y^2-72xy-6x^5+35x^2y^3-18x^6y-32x^3y+33x^3y^2+63y^3+10x^4y+24x^4-14x^2y$$

If we split 3 of the terms as follows:

$$24x^4y^2+28xy^3+30x^5y^2-72xy-6x^5+35x^2y^3-18x^6y+\left(8x^3y-40x^3y \right)+\left(54x^3y^2-21x^3y^2 \right)+63y^3+10x^4y+24x^4+\left(18x^2y-32x^2y \right)$$

Then group as follows:

$$\left(-18x^6y-6x^5+24x^4-21x^3y^2 \right)+\left(30x^5y^2+10x^4y-40x^3y+35x^2y^3 \right)+\left(24x^4y^2+8x^3y-32x^2y+28xy^3 \right)+\left(54x^3y^2+18x^2y-72xy+63y^3 \right)$$

Factor each group:

$$-3x^3\left(6x^3y+2x^2-8x+7y^2 \right)+5x^2y\left(6x^3y+2x^2-8x+7y^2 \right)+4xy\left(6x^3y+2x^2-8x+7y^2 \right)+9y\left(6x^3y+2x^2-8x+7y^2 \right)$$

Factor out common factor:

$$\left(6x^3y+2x^2-8x+7y^2 \right)\left(5x^2y+4xy+9y-3x^3 \right)$$
 
  • #13
how did you know that there are three terms that needs to be split? and why those terms?

did you do trial and error? :confused:
 
  • #14
bergausstein said:
how did you know that there are three terms that needs to be split? and why those terms?

did you do trial and error? :confused:

I did as I suggested you do in an earlier post. I expanded the factorized form to find a possible method of factoring.

Factoring a huge expression like that without knowing the final form would have taken some trial and error on my part.
 
  • #15
i wonder how did you arrive at splitting those terms before getting the factored form.$\displaystyle -\left(3x^3-5x^2y-4xy-9y \right)\left(6x^3y+2x^2-8x+7y^2 \right)$

from this factored form I can get insight of method on how to factor it. but prior to getting this factored form, how did you decide that those three terms should be cut into pieces?
 
  • #16
bergausstein said:
our instructor gave us this problem. i don't know from where did he get this.

i want to learn how to solve this. I've used all the method i know to solve this but no success.

please help me.

paulmdrdo said:
would you help me with this

http://mathhelpboards.com/pre-algebra-algebra-2/challenging-factorization-8267.html

please help me. thanks!

Hello.

Another thing that the brute force I can't think.(Headbang)

If the question, allows you to easily factor with respect to the "x" (6), we can divide, at least in two factors.

1º) We ordered:

x^6(-18y)+x^5(30y^2-6)+x^4(24y^2+10y+24)+x^3(33y^2-32y)+x^2(35y^3-14y)+x(28y^3-72y)+63y^3

2º)
(ax^3+bx^2+cx+d)(ex^3+fx^2+gx+h)=

=x^6(ae)+x^5(af+be)+x^4(ag+bf+ce)+x^3(ah+bg+cf+ed)+x^2(bh+cg+df)+x(ch+dg)+dh

without losing generality, we can assume:

d=7 \ and \ h=9

at the end there is to find the component "y".

x(ch+dg)=x(9c+7g)=x(28y^3-72)

deducting:

c=-8 \ and \ g=4

x^2(bh+cg+df)=x^2(9b-32+7f)=x^2((35y^3-14y)

If \ f=-2 \ then \ b \cancel{\in}{Z}

f=5 \ and \ b=2 \ , \ (9b-32=-14)

And so on. You, see if the system works should try.

Regards.
 
  • #17
bergausstein said:
i wonder how did you arrive at splitting those terms before getting the factored form.$\displaystyle -\left(3x^3-5x^2y-4xy-9y \right)\left(6x^3y+2x^2-8x+7y^2 \right)$

from this factored form I can get insight of method on how to factor it. but prior to getting this factored form, how did you decide that those three terms should be cut into pieces?

I did not do anything before I had the factored form. It was from the factored form that I saw how rewrite the expression, so that I could demonstrate to you a possible path to take.
 
  • #18
how did you get the factored form? :confused: did you use some math software to get that?
 
  • #19
bergausstein said:
how did you get the factored form? :confused: did you use some math software to get that?

Yes, when I referred to W|A, this is wolframalpha.com.
 
  • #20
OH men! I'm scratching my head for hours wondering how did you easily get that factored form. now I understand. I feel bad about myself that I'm not able get the right factored form.

but at any rate, is there a method to do it manually?
 
  • #21
bergausstein said:
OH men! I'm scratching my head for hours wondering how did you easily get that factored form. now I understand. I feel bad about myself that I'm not able get the right factored form.

but at any rate, is there a method to do it manually?

Yes, lots of paper and time for trial and error. :D
 
  • #22
Not necessarily, but as I have mentioned before, it uses far more advanced tools than just pre-algebra. For example, http://www.sigsam.org/bulletin/articles/151/1inaba.pdf paper presents a neat method, although it is mostly impossible to carry the process off-hand.
 

Similar threads

Back
Top