Need help factoring these polynomials?

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    Factoring Polynomial
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Discussion Overview

The discussion revolves around factoring two polynomials: $\displaystyle 2x^2-4xy+2y^2+5x-3-5y$ and $6x^2-xy+23x-2y^2-6y+20$. Participants explore various methods and approaches to factor these expressions, sharing their attempts and seeking guidance on specific steps.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents the first polynomial and their attempt at factoring it, expressing uncertainty about how to proceed.
  • Another participant suggests a factorization for a related polynomial $2u^2+5u-3$ and proposes a connection to the original polynomials.
  • A different participant discusses grouping terms in the second polynomial and proposes a method to factor it, introducing variables $a$ and $b$ to relate to the coefficients.
  • Further replies explore the implications of choosing positive or negative values for $a$ and $b$, questioning the generality of the proposed method.
  • Participants discuss the form of the factorization and its relation to multiplying trinomials, seeking a more generalized expression.

Areas of Agreement / Disagreement

There is no consensus on the best method for factoring the polynomials, as participants explore different approaches and express varying degrees of certainty about their steps and results.

Contextual Notes

Participants express uncertainty regarding the signs of coefficients in their proposed factorizations and the formal naming of the methods used. The discussion includes attempts that may not lead to definitive conclusions, reflecting the exploratory nature of the topic.

Who May Find This Useful

Individuals interested in polynomial factoring techniques, students seeking help with homework, and those exploring mathematical reasoning in algebra may find this discussion beneficial.

bergausstein
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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!
 
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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!
 

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