# Factoring a Polynomial

#### bergausstein

##### Active member
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
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
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
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
Re: factoring

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

what's the next step?

#### MarkFL

##### Administrator
Staff member
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
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
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
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
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.

#### paulmdrdo

##### Active member
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
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
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
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
that's enlightening! ! thanks!