Factor Completely (PreCalculus)

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In summary, the child is trying to do a problem on homework that they do not understand and is lost. They have found the factors of the problem and have found that it is a three by five problem.
  • #1
GreenPrint
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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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  • #2
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)
 
  • #3
wait sorry

(x^3 - 2x^2 + 4x + 3)/( x^2 (x + 1)(x- 1) )
 
  • #4
this is actually the correct answer sorry
5(x + 3)(x - 2)^2 (x + 1)
i was looking at the wrong problem
 
  • #5
wow I'm really losing it
6(3x - 5)(2x + 1)^2 (5x - 4)
 
  • #6
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]
 
  • #7
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
 

1. What does it mean to factor completely?

Factoring completely in precalculus refers to the process of breaking down a polynomial expression into its simplest form. This involves finding the greatest common factor and then using techniques such as grouping, difference of squares, and trinomial factoring to fully factor the expression.

2. Why is factoring important in precalculus?

Factoring is important in precalculus because it allows us to simplify complex polynomial expressions and solve equations. It is also a fundamental skill that is necessary for understanding more advanced concepts in algebra and calculus.

3. What are the steps for factoring completely?

The steps for factoring completely include:

  • Finding the greatest common factor (GCF)
  • Using the difference of squares formula (a² - b² = (a+b)(a-b))
  • Using the trinomial factoring method (ax² + bx + c = (mx + p)(nx + q))
  • Grouping terms with a common factor
  • Checking for any remaining factors that can be factored further

4. How do I know when an expression is completely factored?

An expression is completely factored when it cannot be broken down any further using the techniques mentioned in the previous answer. This means that there are no common factors left and all binomials and trinomials are in their simplest form.

5. What are some common mistakes to avoid when factoring completely?

Some common mistakes to avoid when factoring completely include:

  • Forgetting to check for the GCF before using other factoring techniques
  • Misapplying the difference of squares formula (remember that both terms must be perfect squares)
  • Making errors when grouping terms or factoring trinomials
  • Forgetting to check for any remaining factors that can be factored further

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