Question

my attachment is below!

my answer: x + 4/ 6 (x - 2)

Answer 1

answer: (12 x-96)/((3 x+3) (x^2-4 x-32) (6 x^2-6 x-12))

Answer 2

General strategy for solving this particular kind of problem.

1. Invert and multiply:
\[\huge \dfrac{12x-96}{x^2-4x-32} \cdot \frac{3x+3}{6x^2-6x-12}\]

2. Factor each one of the numerators and denominators separately (four steps)

a\[\huge \dfrac{12(x-8)}{x^2-4x-32} \cdot \frac{3x+3}{6x^2-6x-12}\]
b\[\huge \dfrac{12(x-8)}{(x-8)(x+4)} \cdot \frac{3x+3}{6x^2-6x-12}\]
c\[\huge \dfrac{12(x-8)}{(x-8)(x+4)} \cdot \frac{3(x+1)}{6x^2-6x-12}\]
di (factor out the six)\[\huge \dfrac{12(x-8)}{(x-8)(x+4)} \cdot \frac{3(x+1)}{6(x^2-x-2)}\]
dii (factor the quadratic)\[\huge \dfrac{12(x-8)}{(x-8)(x+4)} \cdot \frac{3(x+1)}{6(x-2)(x+1)}\]

3. Go cancel crazy!

a 12/6 = 2
\[\huge \dfrac{2(x-8)}{(x-8)(x+4)} \cdot \frac{3(x+1)}{(x-2)(x+1)}\]
b get rid of the (x-8)
\[\huge \dfrac{2}{(x+4)} \cdot \frac{3(x+1)}{(x-2)(x+1)}\]
c get rid of the (x+1)
\[\huge \dfrac{2}{(x+4)} \cdot \frac{3}{(x-2)}\]
d multiply 2*3 = 6
\[\huge \dfrac{6}{(x+4)(x-2)} \]

Hope this helps. The ultimate idea here is to break it apart bit by bit. "Divide and conquer". :)