<s>[INST] 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.
 [/INST] 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}
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