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Hard factoring

paulmdrdo

Active member
May 13, 2013
386
this stunned me. i stare at these problem for a while and tried my best to factor it but no success. please help me solve this.

1. $x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x$

2. $3x^2+7xy-3xz-2yz+4y^2-6z^2$
 
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kaliprasad

Well-known member
Mar 31, 2013
1,309
this stunned me. i stare at these problem for a while and tried my best to factor it but no success. please help me solve this.

1. $x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x$

2. $3x^2+7xy-3xz-2yz+4y^2-6z^2$
Not complete solution but outline of solution
put in descending power of x
x^3 + x^2(y^2+1) + x(y-3y^2 – 12) + y(y^2+4)
as y^2 + 4 cannot be factored further it shall be product of a linear and quadratic in x
so (x+/-(y^2 + 4))(x^2 +/- y+ cx) - not pssible as it gives x^2y^2


or (x+/-y))(x^2 + cx +/- (y^2+4)

where c does not conatain x but may contain y

now you can take (x+y)(x^2 + cx + y^2 + 4)

and (x-y)(x^2 + cx - y^2 - 4) and one of them is the result
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
this stunned me. i stare at these problem for a while and tried my best to factor it but no success. please help me solve this.

1. $x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x$
I would look at a factorization of the form:

\(\displaystyle \left(x^2+y+ax \right)\left(x+y^2+b \right)\)

Expand, and then compare coefficients.

2. $3x^2+7xy-3xz-2yz+4y^2-6z^2$
Here I would try the form:

\(\displaystyle (ax+by+cz)(dx+ey+fz)\)
 

paulmdrdo

Active member
May 13, 2013
386
$\displaystyle \left(x^2+y+ax \right)\left(x+y^2+b \right)$ -->expanding this i would get,

$x^3+x^2y^2+bx^2+xy+y^3+by+ax^2+axy^2+abx$

then what's next?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\displaystyle \left(x^2+y+ax \right)\left(x+y^2+b \right)$ -->expanding this i would get,

$x^3+x^2y^2+bx^2+xy+y^3+by+ax^2+axy^2+abx$

then what's next?
Equate that to the original polynomial and equate corresponding coefficients.
 

paulmdrdo

Active member
May 13, 2013
386
what do you mean to equate? set them equal to each other?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
what do you mean to equate? set them equal to each other?
Yes, and then set the corresponding coefficients equal to one another, and solve the resulting simultaneous system.
 

paulmdrdo

Active member
May 13, 2013
386
you mean like this,

$x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x=x^3+x^2y^2+bx^2+xy+y^3+by+ax^2+axy^2+abx$

what's next after this?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
you mean like this,

$x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x=x^3+x^2y^2+bx^2+xy+y^3+by+ax^2+axy^2+abx$

what's next after this?
I would subtract terms common to both sides:

\(\displaystyle x^2+4y-3xy^2-12x=(a+b)x^2+by+axy^2+abx\)

Now, can you find $a$ and $b$?
 

paulmdrdo

Active member
May 13, 2013
386
$a=-3, b=4$

what's next?

how did you know that $x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x$ would factor to this form $\displaystyle \left(x^2+y+ax \right)\left(x+y^2+b \right)$ are you using trial and error? tell me what the trick is please.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Plug those values into:

\(\displaystyle \left(x^2+y+ax \right)\left(x+y^2+b \right)\)

and you will have successfully factored the polynomial.
 

paulmdrdo

Active member
May 13, 2013
386
yes MARKFL i can do that. but some steps aren't still clear to me.

like how did you decide that $x^3+x^2y^2+x^2+xy+y^3+4y-3xy^2-12x$ would have this factored form $\displaystyle \left(x^2+y+ax \right)\left(x+y^2+b \right)$. i know that if we expand this it would be like the original polynomial, but the point that i need an answer to is how did you, in the first place decide it's going to be in that form? what rule or trick are you using? in what way do you look at the polynomial to come up that some specific technique should be used.

i want to know the trick on how to determine what would the factored form of my original polynomial be. especially when it comes to this messy type of polynomial. do i need to use trial and error of jumbling/grouping the terms then factor it?

please bear with me. i find your steps interesting that's why i want to fully comprehend and adopt it. thanks!:)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I noticed that:

\(\displaystyle \left(x^2+y \right)\left(x+y^2 \right)=x^3+x^2y^2+xy+y^3\)

This leaves us with:

\(\displaystyle x^2+4y-3xy^2-12x\)

So the first two terms could be obtained by adding a constant in the second factor, and the last two terms could be obtained by putting a constant times $x$ in the first factor, given the constant in the second factor. some people prefer to try grouping, but I find this method easier for me.
 

paulmdrdo

Active member
May 13, 2013
386
what if the polynomial is given in this way? what would you do first?

$xy+x^2+y^3+4y-3xy^2-12x+x^3+x^2y^2$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
what if the polynomial is given in this way? what would you do first?

$xy+x^2+y^3+4y-3xy^2-12x+x^3+x^2y^2$
The process would be the same...sometimes you have to tinker a bit to see how you want to group the terms. There are other ways to go about it; we each develop our own style that works best for us.