Practice PSAT Question About Rational Functions This is taken directly from a PSAT Practice Test:  

$f(x)=\dfrac{2x-4}{2x^2+2x-4}$
  A rational function is defined above. Which of the following is an equivalent form that displays values not included in the domain as constants or coefficients?  
A) $f(x)=\dfrac{x-2}{x^2+x-2}$
B) $f(x)=\dfrac{2(x-2)}{2(x+2)(x-1)}$  
C) $f(x)=\dfrac{1}{x+1}$  
D)$f(x)=\dfrac{1}{2x^2}$

My understanding is that this question is asking for the form that includes $-2$, and $1$ as either a coefficient or constant, and is equal to the original equation. Both A and B seem to satisfy this, both of them have $-2$ as a constant, and 1 as a coefficient.  
The answer key in the back says it is B, but I and a few of my teachers thought it was A, since A was a simpler version of B. What are we doing wrong here?

The answer should be (b). The domain does not include -2 and 1.  Answer (a) has "-2" as a constant (for x-2 and x^2+x-2) and "1" as a coefficient (for x^2(="1"*x^2) and x terms). No values other than "-2" and "1" as constants or coefficients are in (a). 
Though (b) has "-2" as a constant for x-2 and "1" as a coefficient for x+2 (="1"*x+2), x-2, and x-1, (b) also has "+2" as a constant for x+"2" and as a coefficient for 2(x+2).
