MHB Need help factoring these polynomials?

AI Thread Summary
The discussion focuses on factoring two polynomials: 2x² - 4xy + 2y² + 5x - 3 - 5y and 6x² - xy + 23x - 2y² - 6y + 20. The first polynomial is partially factored as 2(x - y)² + 5(x - y) - 3, but the user is stuck. For the second polynomial, a method involving grouping and trial and error is used to find coefficients a and b, leading to the factorization (3x - 2y + 4)(2x + y + 5). The conversation also touches on the method of undetermined coefficients and the general form for multiplying trinomials. The participants seek clarity on the steps and reasoning behind the factoring process.
bergausstein
Messages
191
Reaction score
0
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!
 
Mathematics news on Phys.org
Re: factoring

bergausstein said:
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:

$$2u^2+5u-3$$ ?
 
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.
 
Re: factoring

The second polynomial is:

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

I would group as follows:

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

We see the first group may be factored as follows:

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

I would next write:

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

Can you find $(a,b)$?
 
Re: factoring

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

what's the next step?
 
Re: factoring

So this means the polynomial can be written as:

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

Can you proceed? If not, try letting:

$$u=2x+y,\,v=3x-2y$$

and you have:

$$uv+4u+5v+4\cdot5$$
 
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.
 
Re: factoring

Once we have:

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

Then we can look at a factorization of the type:

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

Expanding this, we find:

$$(2x+y)(3x-2y)+a(2x+y)+b(3x-2y)+ab$$

And this lead us to write:

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

$$ab=20$$
 
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!:)
 
  • #10
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
 
  • #11
MarkFl

how did you know that

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

has the factorization of this type

$\displaystyle (3x-2y+a)(2x+y+b)$
 
  • #12
paulmdrdo said:
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.
 
  • #13
is that always the form we get when we multiply two dissimilar trinomial?

can you give me a more generalized form.
 
  • #14
paulmdrdo said:
is that always the form we get when we multiply trinomials?

can you give me a more generalized form.

Consider the expansion of:

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

Notice we have the form:

$$(ax+by+c)(dx+ey+f)=Ax^2+Bxy+Cy^2+Dx+Ey+F$$
 
  • #15
that's enlightening! (Star)(Star)(Star)(Star)(Star)! thanks!
 
Back
Top