f(x) = ((x+3)(x-4))/(x+5)
Note that f(x) is the product of 3 terms:
(x+3), (x-4) and 1/(x+5)
These terms change sign when x=-5, x=-3 and x=4. Hence f(x) changes sign at each of these values of x.
When x=-5 the denominator is zero, so f(x) is undefined. So x=-5 is not part of the solution space.
When x=-3 or x=4 the numerator is 0 and the inequality is satisfied. So the solution space includes { -3, 4 }
The three values of x cut the Real line into 4 intervals in which the sign of f(x) does not change, namely:
(-oo, -5), (-5, -3), (-3, 4), (4, oo)
For large positive values of x we can see that f(x) > 0. So the signs of the left hand side in each of these intervals is:
(-oo, -5)": " f(x) < 0
(-5, -3)": " f(x) > 0
(-3, 4)": " f(x) < 0
(4, oo)": " f(x) > 0
Remembering that { -3, 4 } is part of the solution space, the complete solution space is:
(-5, -3] uu [4, oo)
graph{((x+3)(x-4))/(x+5) [-11.83, 8.17, -30.8, 29.2]}
