Need help factoring these polynomials?

[bergausstein]

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]

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]

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]

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]

Re: factoring

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

what's the next step?

 

[MarkFL]

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]

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]

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]

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]

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

 

[paulmdrdo1]

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]

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.

 

[paulmdrdo1]

is that always the form we get when we multiply two dissimilar trinomial?

can you give me a more generalized form.

 

[MarkFL]

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:

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

 

[paulmdrdo1]

that's enlightening! (Star)(Star)(Star)(Star)(Star)! thanks!