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Challenging factorization

bergausstein

Active member
Jul 30, 2013
191
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:

LATEBLOOMER

New member
Aug 7, 2013
21
no one will help you. because it's tedious and impossible. :p
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Are you sure you've copied the expression correctly?
 

bergausstein

Active member
Jul 30, 2013
191
uhm oh. there's a typo. $24x^4y^2$ should be $24x^4$ sorry. i edited now.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

bergausstein

Active member
Jul 30, 2013
191
I'm really sorry. i guess i edited it correctly this time. please bear with me.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

I suggest looking at expanding that to gain insight into how it factors.
 

bergausstein

Active member
Jul 30, 2013
191
yes this the answer. but I'm having a hard time understanding how did you arrive at the answer. please help. anybody?
 
Last edited:

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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?
 

paulmdrdo

Active member
May 13, 2013
386
this is hard.
 
Last edited:

bergausstein

Active member
Jul 30, 2013
191
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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given to factor:

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

If we split 3 of the terms as follows:

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

Then group as follows:

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

Factor each group:

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

Factor out common factor:

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

bergausstein

Active member
Jul 30, 2013
191
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:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

bergausstein

Active member
Jul 30, 2013
191
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

Well-known member
Nov 29, 2013
172
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:
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:

[tex]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[/tex]

2º)
[tex](ax^3+bx^2+cx+d)(ex^3+fx^2+gx+h)=[/tex]

[tex]=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[/tex]

without losing generality, we can assume:

[tex]d=7 \ and \ h=9[/tex]

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

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

deducting:

[tex]c=-8 \ and \ g=4[/tex]

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

[tex]If \ f=-2 \ then \ b \cancel{\in}{Z}[/tex]

[tex]f=5 \ and \ b=2 \ , \ (9b-32=-14)[/tex]

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

Regards.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
how did you get the factored form? :confused: did you use some math software to get that?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

bergausstein

Active member
Jul 30, 2013
191
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
MHB Math Helper
Mar 22, 2013
573
Not necessarily, but as I have mentioned before, it uses far more advanced tools than just pre-algebra. For example, this paper presents a neat method, although it is mostly impossible to carry the process off-hand.