Question 746174
if we invert the divisor and rewrite as multiplication then

{{{
((4x^2-x-12)/(x^2+9x+18)) / ((x^2+x-20)/(x^2+8x+15))=

((4x^2-x-12)/(x^2+9x+18))* ((x^2+8x+15)/(x^2+x-20))
 }}}


Factoring is a little tedious, but you can verify that when factored we get:



{{{

((4x^2-x-12)/(x^2+9x+18))* ((x^2+8x+15)/(x^2+x-20))=

(
(4x^2-x-12)/((x+3)(x+6))
)
*
(
(
(x+3)(x+5)
)
/
((x+5)(x-4))

)
 }}}


next we cancel:



{{{

((4x^2-x-12)/(x^2+9x+18))* ((x^2+8x+15)/(x^2+x-20))=


(4x^2-x-12)/((x+6)(x-4))

 }}}


and finally...


{{{
(4x^2-x-12)/(x^2+2x-24)

 }}}