Factor Completely (PreCalculus)

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Homework Help Overview

The problem involves factoring a polynomial expression completely, specifically the expression 2(3x - 5)3(2x + 1)^3 + (3x - 5)^2 (3)(2x + 1)^2 (2). Participants are exploring the steps necessary to simplify and factor this expression.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rewriting the expression for clarity and identify common factors in the terms. There are attempts to factor out shared components, such as (3x - 5) and (2x + 1).

Discussion Status

Some participants have provided partial factorizations and explored different interpretations of the expression. There is a mix of uncertainty and attempts to clarify the factoring process, with no explicit consensus on the final form yet.

Contextual Notes

Participants express confusion regarding the initial setup of the problem and the correct interpretation of the expression to be factored. There are indications of miscommunication about the problem statement, as some participants reference different problems or answers.

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Homework Statement




Factor completely

2(3x - 5)3(2x + 1)^3 + (3x - 5)^2 (3)(2x + 1)^2 (2)



Homework Equations





The Attempt at a Solution



I'm not really sure how to do this problem

I started out by rewriting this as it seemed to be written all weird like
6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2

I am unsure were do go from here...
 
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this was the answer in the back book. I do not see how to get this though

5(x + 3)(x - 2)^2 (x + 1)
 
wait sorry

(x^3 - 2x^2 + 4x + 3)/( x^2 (x + 1)(x- 1) )
 
this is actually the correct answer sorry
5(x + 3)(x - 2)^2 (x + 1)
i was looking at the wrong problem
 
wow I'm really losing it
6(3x - 5)(2x + 1)^2 (5x - 4)
 
GreenPrint said:

Homework Statement




Factor completely

2(3x - 5)3(2x + 1)^3 + (3x - 5)^2 (3)(2x + 1)^2 (2)
Look at the two separate parts- that are separated by the "+":
2(3x-5)(3)(2x+1)^3 and
(3x-5)^2(3)(2x+1)^2

You see that each has a (3x-5) factor (one squared), a factor of (3), and a factor of (2x+1) (one squared and the other cubed. We can take those out, using the "distributive law" ab+ ac= a(b+ c).

That is, we can factor out (3x-5) and (3) and two copies of 2x+1 since there are at least 2 in each. That gives (3x-5)(3)(2x+1)^2(2(2x+1)+ (3x+5)= 3(3x-5)(2x+1)^2(4x+ 2+ 3x+5)= 3(3x- 5)(2x+1)^2(7x+ 7). And since there is now a "7" in both parts ofthe last factor, we can take that out to get 21(3x-5)(2x+1)^2(x+ 1)


Homework Equations





The Attempt at a Solution



I'm not really sure how to do this problem

I started out by rewriting this as it seemed to be written all weird like
6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2

I am unsure were do go from here...[/QUOTE]
 
GreenPrint said:


I started out by rewriting this as it seemed to be written all weird like
6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2

I am unsure were do go from here...


Collect the common factors in both terms: they are 6, (2x+1)^2, (3x-5).

6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2=
=6(3x-5) (2x+1)^2 (2x+1+3x-5)=
6(3x-5) (2x+1)^2 (5x-4).

ehild
 

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