How do you simplify and find the excluded values of #(v^2+4) / (v^2-3v-18)#?

Excluded values mean asymptotes and holes. So let's look for them:

First, let's expand all our components:

#(v^2+4)/(v^2-3v-18)#

The numerator almost looks like a differnce of squares, but it's adding instead of subtracting. That means that the expanded version of it will have imaginary numbers (#i#) in it. Let's not expand it if we don't have to.

The denominator is easier. We just need to factor.
#v^2-3v-18#

We are looking for two numbers that add to #-3# and multiply to #-18#. We know that one of the numbers will be negative, because to get #color(red)(-)18#, we need to multiply a positive number by a negative number.

To find the factors, let's ignore the signs for now:
# color(white)(xx)+ 3#
# color(white)(+)xx 18#
........................
#1 xx 18#
#2 xx 9#
#color(red)(3 xx 6)#

#3+ -6=-3# and #-6 xx 3=18#

Now we have our factors:

#(v-6)(v+3)#

#(v^2+4)/((v-6)(v+3))#

Asymptotes are values of #v# that eqaul a division by #0#. To solve for them, we set each component in the denominator equal to #0# and solve for #v#:

Case 1

#v-6=0#

#v=6#

Case 2

#v+3=0#

#v=-3#

So, when #v=6# or #-3#, the denominator becomes #0#, creating an asymptote.
there are no holes (no identical factors in the numerator and denominator), so the only excluded values are #v=6 and -3#

Just to check our work, let's graph the equation and see
graph{y=(x^2+4)/((x-6)(x+3))}