Answer
$(x-5)(x+4)(2x-1)$
Work Step by Step
Step 1. Factor both denominators.
The first one is a trinomial $x^{2}+bx+c.$
... We find factors of $c=-20$ whose sum is $b=-1$.
... these are $-5$ and $+4$.
$x^{2}-x-20=(x-5)(x+4)$
The second one is a trinomial $ax^{2}+bx+c.$
... We find factors of $ac=-8$ whose sum is $b=+7$.
... these are $+8$ and $-1$
... rewrite $bx$ and factor in pairs:
$2x^{2}+7x-4=2x^{2}+8x-x-4$
$=2x(x+4)-(x+4)$
$=(x+4)(2x-1)$
Step 2. List the factors of the first denominator.
$ List=(x-5),(x+4),...\qquad$ (list in progress)
Step 3. Add to the list in step 2 any factors of the second denominator that are not yet listed.
$(x+4)$ is already in the list.
$(2x-1)$ is not in the list, we add it to the list
$List=(x-5),(x+4),(2x-1),...$
Step 4. LCD is the product of the listed factors.
$LCD=(x-5)(x+4)(2x-1)$
