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]