Question

Solve the inequality (x^2-3x-4)//(x^2-8x+15)>0

Answer 1

first write as
\[\frac{(x-4) (x+1)}{(x-5) (x-3)}>0\]

Answer 2

changes sign at -1,3,4,5
so you have to consider the intervals
\[(-\infty,-1),(1,3),(3,4),(4,5),(5,\infty)\]

Answer 3

you can test one and then you will have them all because it changes sign on each interval. easiest probably is to test interval
\[(-1,3) \] by replacing x by 0

Answer 4

if x = 0 three of those factors are negative, so the whole thing will be negative, and therefore the interval
\[(-1,3)\] is out
your answer will be
\[(-\infty, -1)\cup (3,4)\cup (5,\infty)\]

Answer 5

first write as (x-4)(x+1)/(x-5)(x-3)>0.. make any factor as 0. and we have x = -1,3,4,5..
so the result of inequality are \[x <-1 \cup 3<x<4 \cup x>5\]