Factoring by grouping: how to deal when the factors extracted are of the form $(x+k)$ and $ (x-k) $. EDITED : the original post contained a mistake regarding the factorization of the numerator.

The source of the exercice I'm trying to do is : Barton's College Practice Placement Test, Q. 17 https://www.barton.edu/pdf/math/practice-math-placement-test.pdf
Question : Simplify  $\frac {4x^2 -1} { 2x^2 +5x -3}$ .
I first tried to factorize the denominator  :
By grouping I get : $2x^2 +5x -3 = (2x^2+2x)+ (3x-3) = 2x (x+1) +3(x-1)$.
My question is : is there a classic way to deal with that kind of situation? I mean a way to recover an $(x+1)$ instead of an $(x-1)$ in the second term ( or reciprocally)?
I know how to get $-( b-a)$ from $(a-b) $  but that wouldn't help here.
 A: For a quadratic $$ax^2+bx+c=a(x-x_1)(x-x_2)$$
Here $x_{1,2}$ are the roots of $ax^2+bx+c=0$:
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
A: One way is to try to get the factors of the numerator into the denominator:
$$4x^2 -1 = (2x-1)(2x+1)$$
and
$$2x^2 +5x -3 = (2x^2 -x) + (6x-3) = x(2x-1) + 3(2x-1) = (2x-1)(x+3).$$
Another possibility (if you do not wish to study the discriminant formula) is to complete the square:
\begin{align}
2x^2 + 5x -3 &= 2\left(x^2 + 2\cdot \frac{5}{4}x - \frac{3}{2}\right)\\
             &= 2\left(x^2 + 2\cdot \frac{5}{4}x + \frac{25}{16} - \frac{25}{16}- \frac{3}{2}\right)\\
             &=2\left(\left(x + \frac{5}{4}\right)^2 - \frac{49}{16}\right)\\ 
             &=2\left(x+ \frac54 + \frac74\right)\left(x + \frac54 - \frac74\right)\\
             &=2\left(x+3\right)\left(x-\frac12\right)
\end{align}
A: This is intended to address your actual question, and not how to solve the problem that inspired your question.
Given that one can simultaneously work with $a-b$ and $b-a$ as you indicated, I think it is natural to wonder whether some similar method could allow you to simultaneously work with $x+1$ and $x-1.$ However, this is not the case. Here's one way of seeing this, although it's certainly NOT something anyone would expect you to have thought about. Indeed, it's often much more difficult to prove you can't do something than to prove you can do something, since in the latter case this can be accomplished by simply doing it.
In order for you to simultaneously work with $x-1$ and $x+1,$ it is necessary (and sufficient, for that matter) that they are nonzero constant multiplies of each other (like $2x+6$ and $x+3$ are, or like $3x-5$ and $5-3x$ are). Maybe this needs some justification for you, maybe not, but if you allow me to assume this, then here's why $x+1$ and $x-1$ are not nonzero multiples of each other. (They obviously don't appear to be nonzero multiples of each other, but how do we absolutely know for sure? That's what I'll show.)
In order that $x+1$ and $x-1$ are nonzero multiples of each other, there must be a nonzero number $c$ such that $x+1 = c(x-1).$ (See Technical Note below.) Since the same value of $c$ is supposed to work for all values of $x,$ we should get the same value of $c$ when $x=2$ as we do when $x=3.$ When $x=2,$ the equation $x+1 = c(x-1)$ becomes $2+1 = c(2-1),$ so $c = 3.$ However, when $x=3,$ the equation $x+1 = c(x-1)$ becomes $3+1 = c(3-1),$ or $4 = 2c,$ so $c = 2.$ Since we don't get the same value of $c$ in these two cases, it is NOT TRUE that there is a nonzero value of $c$ such that $x+1 = c(x-1)$ (i.e. it is not possible for the same $c$ to work for all values of $x).$ Therefore, it is NOT TRUE that $x+1$ and $x-1$ are nonzero multiples of each other.
Technical Note: If we're being nit-picks about this, then $x+1 = c(x-1)$ is the actually the condition that $x+1$ is a nonzero multiple of $x-1.$ However, since this last equation implies $x-1 = \frac{1}{c}(x+1),$ we also have that $x-1$ is a nonzero multiple of $x+1.$ So $x+1$ being a nonzero multiple of $x-1$ implies that $x-1$ is a nonzero multiple of $x+1.$ In a similar way, one can show that $x-1$ being a nonzero multiple of $x+1$ implies that $x+1$ is a nonzero multiple of $x-1.$ Therefore, the condition that either is a nonzero multiple of the other is equivalent to the condition that each is a nonzero multiple of the other, and so saying they are nonzero multiples of each other (without saying which is the nonzero multiple of which) is not ambiguous, and is equivalent to the specific condition that $x+1$ is a nonzero multiple of $x-1.$ (Whew!)
