Factor completely
2(3x - 5)3(2x + 1)^3 + (3x - 5)^2 (3)(2x + 1)^2 (2)
I'm not really sure how to do this problem
I started out by rewriting this
6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2
I am unsure were do go from here...
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)
6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2
6(2x+1)^2(2x+1)(3x-5) +6(2x+1)^2 (3x-5)(3x-5)
6(2x+1)^2(3x-5) [ 2x+1 + 3x-5 ]
6(2x+1)^2(3x-5)(5x-4)
So I do not see how to get that answer either.
To factor completely, you need to look for common factors among the terms and express the expression as a product of those factors. Let's start by factoring out the common factor, which in this case is (2x + 1) and (3x - 5).
First, let's look at the first term, 6(2x + 1)^3(3x -5). We can factor out (2x + 1) from this term:
6(2x + 1)^3(3x -5) = 6(2x + 1)(2x + 1)(2x + 1)(3x -5)
Next, let's look at the second term, 6(3x - 5)^2(2x + 1)^2. We can factor out (3x - 5) and (2x + 1) from this term:
6(3x - 5)^2(2x + 1)^2 = 6(3x - 5)(3x - 5)(2x + 1)(2x + 1)
Now, we have factored both terms into their common factors:
6(2x + 1)(2x + 1)(2x + 1)(3x - 5) + 6(3x - 5)(3x - 5)(2x + 1)(2x + 1)
To simplify this expression further, notice that we have common factors (2x + 1) and (3x - 5) in both terms. We can factor them out:
(2x + 1)(3x - 5)[6(2x + 1)(2x + 1)(2x + 1) + 6(3x - 5)(3x - 5)(2x + 1)(2x + 1)]
Now, we have factored the expression completely. The expression can be written as the product of (2x + 1)(3x - 5) multiplied by the remaining terms:
(2x + 1)(3x - 5)[6(2x + 1)(2x + 1)(2x + 1) + 6(3x - 5)(3x - 5)(2x + 1)(2x + 1)]