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Factoring a Polynomial

bergausstein

Active member
Jul 30, 2013
191
can you guys help me factor this polynomial.

$\displaystyle 2x^2-4xy+2y^2+5x-3-5y$

$6x^2-xy+23x-2y^2-6y+20$

by the way this is what i tried in prob 1

$2(x-y)^2+5(x-y)-3$ -->> i'm stuck here. and in prob 2 i have no idea where and how to start.

thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: factoring

can you guys help me factor this polynomial.

$\displaystyle 2x^2-4xy+2y^2+5x-3-5y$

$6x^2-xy+23x-2y^2-6y+20$

by the way this is what i tried in prob 1

$2(x-y)^2+5(x-y)-3$ -->> i'm stuck here. and in prob 2 i have no idea where and how to start.

thanks!
How would you factor:

\(\displaystyle 2u^2+5u-3\) ?
 

bergausstein

Active member
Jul 30, 2013
191
Re: factoring

oh yes! that rings a bell!

$(2u-1)(u+3)=(2x-2y-1)(x-y+3)$ -->>>this should be the answer.

can you also give me hints on prob 2.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: factoring

The second polynomial is:

\(\displaystyle 6x^2-xy+23x-2y^2-6y+20\)

I would group as follows:

\(\displaystyle \left(6x^2-xy-2y^2 \right)+\left(23x-6y \right)+20\)

We see the first group may be factored as follows:

\(\displaystyle (2x+y)(3x-2y)+\left(23x-6y \right)+20\)

I would next write:

\(\displaystyle a(2x+y)+b(3x-2y)=23x-6y\) where \(\displaystyle ab=20\)

Can you find $(a,b)$?
 

bergausstein

Active member
Jul 30, 2013
191
Re: factoring

by using trial and error i get a=4, b=5.

what's the next step?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: factoring

So this means the polynomial can be written as:

\(\displaystyle (2x+y)(3x-2y)+4(2x+y)+5(3x-2y)+4\cdot5\)

Can you proceed? If not, try letting:

\(\displaystyle u=2x+y,\,v=3x-2y\)

and you have:

\(\displaystyle uv+4u+5v+4\cdot5\)
 

bergausstein

Active member
Jul 30, 2013
191
Re: factoring

yes, the answer is $(3x-2y+4)(2x+y+5)$.

Markfl can you tell me what rule do you have in mind to come up with this:

$\displaystyle a(2x+y)+b(3x-2y)=23x-6y$ where $ab=20$

i want to fully understand the steps you showed me in prob 2.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: factoring

Once we have:

\(\displaystyle (2x+y)(3x-2y)+\left(23x-6y \right)+20\)

Then we can look at a factorization of the type:

\(\displaystyle (3x-2y+a)(2x+y+b)\)

Expanding this, we find:

\(\displaystyle (2x+y)(3x-2y)+a(2x+y)+b(3x-2y)+ab\)

And this lead us to write:

\(\displaystyle a(2x+y)+b(3x-2y)=23x-6y\)

\(\displaystyle ab=20\)
 

bergausstein

Active member
Jul 30, 2013
191
Re: factoring

should the sign preceeding a and b always positve? or it depends?

$\displaystyle (3x-2y+a)(2x+y+b)$

and what's the name for this method of factoring? or how would you name it at least?;)

thanks! you're such a help!:)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: factoring

I chose positive signs, but $a$ and/or $b$ may be negative. I don't think this method has a formal name. We may choose to call it the method of undetermined coefficients, to borrow a term from solving certain differential equations. :D
 

paulmdrdo

Active member
May 13, 2013
386
MarkFl

how did you know that

$\displaystyle (2x+y)(3x-2y)+\left(23x-6y \right)+20$

has the factorization of this type

$\displaystyle (3x-2y+a)(2x+y+b)$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
MarkFl

how did you know that

$\displaystyle (2x+y)(3x-2y)+\left(23x-6y \right)+20$

has the factorization of this type

$\displaystyle (3x-2y+a)(2x+y+b)$
I didn't know it would actually factor that way, but it seemed to be the best form to try.
 

paulmdrdo

Active member
May 13, 2013
386
is that always the form we get when we multiply two dissimilar trinomial?

can you give me a more generalized form.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
is that always the form we get when we multiply trinomials?

can you give me a more generalized form.
Consider the expansion of:

\(\displaystyle (ax+by+c)(dx+ey+f)=adx^2+(ae+bd)xy+bey^2+(af+cd)x+(bf+ce)y+cf\)

Notice we have the form:

\(\displaystyle (ax+by+c)(dx+ey+f)=Ax^2+Bxy+Cy^2+Dx+Ey+F\)
 

paulmdrdo

Active member
May 13, 2013
386
that's enlightening! (Star)(Star)(Star)(Star)(Star)! thanks!