Divide the rational expressions below and simplify as much as possible: The question is : 

$$\frac{x^2 + 8}{2x^2 + x - 3}\div\frac{x + 2}{x - 1}$$

the answer I get is 
$$\frac{x^2 + 8}{(2x + 3)(x + 2)}$$
but a further factor of $x+2$ is factored out somehow from the $x^2+8$. 
I'm having doubts about this. please help, thanks 
ANSWER GIVEN BY PROF:

$$\frac{(x+2)\left(x^{2}-2 x+4\right)}{(x-1)(2 x+3)} \cdot \frac{x-1}{x+2}=\frac{x^{2}-2 x+4}{2 x+3}$$


In the answer given by the prof $(x+2)(x^2-2x+4)=x^3+8$.  It seems the problem statement has the wrong exponent on the $x$ at the start.  Your answer is correct for the problem as given.
