Factoring Problem. It is confusing and i need it for tomorrow.?

Question

(3/ x^2 - x - 2) - (4/x^2 +x -6) = 1/ 3x+3

Answer 1

(3/ x^2 - x - 2) - (4/x^2 +x -6) = 1/ 3x+3
=3((x-2)(x+1) -4/((x+3)(x-2)) = 1/3(x+1)
So your common denominator is 3(x+3)(x+1)(x-2)
3*3(x+3) -3*4(x+1) = (x+3)(x-2)
9x +27 -12x -12 =x^2 +x -6
-3x +15 = x^2 +x -6
x^2 +4x -21=0
(x+7)(x-3) =0
x = -7 and x = 3

Answer 2

Reading question as:-
3 / (x² - x - 2) - 4 / (x² + x - 6) = 1 / (3x + 3)
3 / (x - 2) (x + 1) - 4 / (x + 3) (x - 2) = 1 / 3 (x + 1)

3 / (x - 2) - 4 (x + 1) / (x + 3) (x - 2) = 1 / 3
3 - 4 (x + 1) / (x + 3) = (x - 2) / 3
3 (x + 3) - 4 (x + 1) = (x - 2) (x + 3) / 3
9 (x + 3) - 12 (x + 1) = (x - 2) (x + 3)
9x + 27 - 12 x - 12 = x² + x - 6
15 - 3x = x² + x - 6
x² + 4x - 21 = 0
(x + 7) (x - 3) = 0
x = - 7 , x = 3

Answer 3

It's hard to tell the equation since you don't use many parentheses, especially the last term.
Assuming the trinomials are all in the denominator, you would factor
x^2 - x - 2 = (x-2)(x+1) and
x^2 + x - 6 = (x-2)(x+3) and
3x + 3 = 3 (x+1)
The least common denominator would then be:
(x-2)(x+1)(x+3)
Multiply both sides of the equation by the LCD and cancel terms until you get:
3(x+3) - 4(x+1) = (1/3)(x-2)(x+3)
At this point, FOIL the right hand side, collect all the terms on one side of the equation and you will have a quadratic. Either factor this or use the quadratic formula and you are done!