Assuming that x is restricted to the set of real numbers, we can't take square roots of negative numbers. So, -x3+5x2+9x-45 ≥0.
Factor by grouping: -x2(x-5) + 9(x-5) ≥ 0
(x-5)(-x2+9) ≥ 0
(x-5)(-x+3)(x+3) ≥ 0
The left side is equal to zero when x = 5, 3, or -3.
_________l___________l____________l__________
-3 3 5
Choose a test point from each of the intervals (-∞,-3), (-3,3), (3,5), (5,∞) and plug into the factored form above.
Interval Test Point Sign of (x-5)(-x+3)(x+3)
(-∞, -3) -4 positive
(-3,3) 0 negative
(3,5) 4 positive
(5,∞) 6 negative
Since we are trying to solve the inequality (x-5)(-x+3)(x+3)≥0, we choose the intervals in the table above for which we got "positive" in the rightmost column. We also include the endpoints (not -∞) as part of the solution set since, at those points, (x-5)(-x+3)(x+3) = 0.
Solution set: (-∞, -3] ∪ [3,5]