SOLUTION: I am having trouble solving this, I have tried it up to the point where I need get the final answer but I have a feeling its wrong, Can you please check to see what I am doing wron

Question 36479This question is from textbook
: I am having trouble solving this, I have tried it up to the point where I need get the final answer but I have a feeling its wrong, Can you please check to see what I am doing wrong and help out

Divide:
2x^2 + 5x � 12/9x^2 � 16 � 2x^2 � 7x + 6/3x^2 � x � 4

I inverted the divisor
2x^2 + 5x - 12/ 9x^2 - 16 � 3x^2 - x - 4/2x^2 - 7x + 6

Now I am going to set it up to multiply the numerators and denominators
(2x^2 + 5x - 12)(3x^2 - x - 4)
(9x^2 - 16)(2x^2 - 7x + 6)
Now I am going to factor
(2x - 3)(x + 4) (3x^2 - x -4)
(3x + 4)(3x - 4)(2x - 3)(2x - 2)

I can only see one thing that will cancel out and that is the (2x - 3)
but I dont think the answer is
(x + 4)(3x^2 - x - 4)
(3x + 4)(3x - 4)(2x - 2)
I think I have missed a step here or I did not factor correctly, please help, I am having a hard time with this one.
Thank you very much!

This question is from textbook

Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I see two factoring errors. First the factors into . Second, where you have a factor of (2x-2), I think it should be (x-2).

In your last step you have this:
(x + 4)(3x^2 - x - 4)
(3x + 4)(3x - 4)(2x - 2)

I think it should be this:
(x + 4)(3x-4)(x+1)
(3x + 4)(3x - 4)(x - 2)

Then the final answer should be

R^2 at SCC

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