Consider the family of functions of domain $\mathbb{R}\setminus \{a\}$, $\frac{2x^2+x-3}{x-a}$.Find a so that there are no vertical asymptotes 
Consider the family of functions of domain  $\mathbb{R}\setminus
 \{a\}$ defined by $f(x) = \frac{2x^2+x-3}{x-a}$.
Determine $a$ so that $f$ has no vertical
  asymptotes.

I know that one way to solve this is by factoring ${2x^2+x-3}$ into $(x-a)$ times something. 
I'm not very good at factoring so I messed around a bit to see if I could find the solution.
My book says the solutions are x = 1 and x = -1.5. I looked at the graph of the function an realised these were the zeros of $2x^2+x-3$.
With a little help from Wolfram I found out how to factor $2x^2+x-3$ : $(x-1)(2x+3)$.
And so $a = 1$
To get the other solution I tried this:
$${2x^2+x-3} = (x-a)(y+z)$$
$x \cdot y = 2x^2 \Leftrightarrow y = 2x$
$${2x^2+x-3} = (x-a)(2x+z) = 2x^2+xz-2xa-az = 2x^2+x(z-2a) -az$$
$az = 3$
$x(z-2a) = x \Leftrightarrow z-2a = 1$
Then I put the above two conclusions into a system and calculated them and got $a = 1 \lor a = -1.5$, no problem there.
My questions:


*

*Why does $a$ take the values of the zeros of $2x^2+x-3$? Is it just a coincidence?


*Is there a simpler way of solving this?



Note that $$f(x) = \frac{2x^2+x-3}{x-a} = \frac{(2x+3)(x-1)}{x-a} = \frac{2(x+1.5)(x-1)}{x-a}.$$ Thus if $a=-1.5$ or $a=1$, then $f(x)$ simplifies to a linear polynomial over the domain $\mathbb{R} \setminus \{a\}$, and linear polynomials do not have vertical asymptotes. 
