Practice PSAT Question About Rational Functions

Question

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?

Answer 1

Factoring the denominator, as in (b), displays the values not included in the domain plainly. It's clear by inspection that $-2, 1$ aren't in the domain when looking at choice (b), whereas it's not in (a).