Question 255649
The problem is
(i) {{{((b-1)/(b+6))}}} / {{{((b-3)/(b^2+10b+24))}}}
factor b^2 + 10b + 24 = (b+6)(b+4), so we get
(ii) {{{((b-1)/(b+6))}}} / {{{((b-3)/((b+6)(b+4)))}}}
we have a couple of restrictions where
b+6=0
b-3=0
b+4=0.
So b cannot = -6, 3, -4.
we can re express (ii) as
(iii) {{{((b-1)/(b+6))}}} * {{{((b+6)(b+4)/(b-3))}}}
and canceling, we get
(iv) {{{(b-1)}}} * {{{(b+4)/(b-3)}}}
which is
(v) {{{(b^2+3b-4)/(b-3)}}}