# Challenging factorization

#### bergausstein

##### Active member
factor

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

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#### MarkFL

Staff member
Are you sure you've copied the expression correctly?

#### bergausstein

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

#### MarkFL

Staff member
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

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

#### MarkFL

Staff member
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.

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#### mathbalarka

##### Well-known member
MHB Math Helper
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?

this is hard.

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#### bergausstein

##### Active member
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.

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#### MarkFL

Staff member
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
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

Staff member
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

##### Active member
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
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.

paulmdrdo said:
Hello.

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

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

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

#### MarkFL

Staff member
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

##### Active member
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?