Homework Help: Factor Completely (PreCalculus)

1. Jan 5, 2010

GreenPrint

1. The problem statement, all variables and given/known data

Factor completely

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

2. Relevant equations

3. 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...

2. Jan 5, 2010

GreenPrint

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. Jan 5, 2010

GreenPrint

wait sorry

(x^3 - 2x^2 + 4x + 3)/( x^2 (x + 1)(x- 1) )

4. Jan 5, 2010

GreenPrint

this is actually the correct answer sorry
5(x + 3)(x - 2)^2 (x + 1)
i was looking at the wrong problem

5. Jan 5, 2010

GreenPrint

wow i'm really losing it
6(3x - 5)(2x + 1)^2 (5x - 4)

6. Jan 5, 2010

HallsofIvy

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)

2. Relevant equations

3. 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. Jan 6, 2010

ehild

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