Unraveling the Complexities of Factorization in Multivariable Polynomials

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Discussion Overview

The discussion revolves around the factorization of a complex multivariable polynomial. Participants explore various methods and challenges associated with factoring the expression, seeking understanding and assistance with the problem presented by a user.

Discussion Character

  • Homework-related
  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about how to start factoring the polynomial.
  • Another participant humorously suggests that the task is tedious and impossible.
  • There is a correction regarding a typo in the polynomial, which leads to a revised expression.
  • A participant shares a factored form obtained from Wolfram Alpha, suggesting that expanding it could provide insight into the factorization process.
  • Some participants express difficulty in understanding how the factored form was derived and inquire about the methods used.
  • One participant mentions the lack of obvious factorization techniques and the complexity of polynomial factorization algorithms.
  • Another participant describes a method involving grouping and splitting terms to facilitate factorization.
  • There are questions about the reasoning behind choosing specific terms to split and whether trial and error was involved in finding the factorization.
  • Some participants discuss the use of mathematical software to obtain the factored form, while others express a desire to learn manual methods for factorization.
  • One participant mentions that advanced tools are often required for such complex factorizations, indicating that it may not be feasible to do manually.

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the methods for factorization, with multiple competing views on how to approach the problem and varying levels of understanding expressed throughout the discussion.

Contextual Notes

Some participants note the limitations of their current knowledge and the complexity of the polynomial, indicating that the problem may require advanced techniques beyond pre-algebra.

bergausstein
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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.
 
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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?
 
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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.
 
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  • #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.
 
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  • #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.
 

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