Unraveling the Complexities of Factorization in Multivariable Polynomials

[bergausstein]

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.

 

[paulmdrdo]

no one will help you. because it's tedious and impossible. :p

 

[MarkFL]

Are you sure you've copied the expression correctly?

 

[bergausstein]

uhm oh. there's a typo. $24x^4y^2$ should be $24x^4$ sorry. i edited now.

 

[MarkFL]

uhm oh. there's a typo. $24x^4y^2$ should be $24x^4$ sorry. i edited now.

W|A still finds no factored form.

 

[bergausstein]

I'm really sorry. i guess i edited it correctly this time. please bear with me.

 

[MarkFL]

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.

 

[bergausstein]

yes this the answer. but I'm having a hard time understanding how did you arrive at the answer. please help. anybody?

 

[mathbalarka]

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?

 

[paulmdrdo1]

this is hard.

 

[bergausstein]

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.

 

[MarkFL]

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:

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

 

[bergausstein]

how did you know that there are three terms that needs to be split? and why those terms?

did you do trial and error?

 

[MarkFL]

how did you know that there are three terms that needs to be split? and why those terms?

did you do trial and error?

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.

 

[bergausstein]

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?

 

[mente oscura]

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.

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.

 

[MarkFL]

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.

 

[bergausstein]

how did you get the factored form? did you use some math software to get that?

 

[MarkFL]

how did you get the factored form? did you use some math software to get that?

Yes, when I referred to W|A, this is wolframalpha.com.

 

[bergausstein]

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?

 

[MarkFL]

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

 

[mathbalarka]

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.