(x + 2)(x − 1)(x − 5) < 0
excludedincluded / zeropoledomain edgesolution/trail
f(-4) =
-90 → satisfiesspeed0.6×
interval(−∞, -2) ∪ (1, 5)
set-builder{ x ∈ ℝ : f(x) < 0 }
Templates & parameters
f(x) = (x − r₁)(x − r₂)(x − r₃)
root r₁−2
-8-208
root r₂+1
-8018
root r₃+5
-8058
Sign chart
| x < -2 | -2 | -2 – 1 | 1 | 1 – 5 | 5 | x > 5 | |
|---|---|---|---|---|---|---|---|
| (x + 2) | − | + | + | + | |||
| (x − 1) | − | − | + | + | |||
| (x − 5) | − | − | − | + | |||
| f(x) | − | + | − | + |
Explanation
at x = -4
| (x + 2) | = ((-4) + 2) | = -2 | − |
| (x − 1) | = ((-4) − 1) | = -5 | − |
| (x − 5) | = ((-4) − 5) | = -9 | − |
product of signs: − · − · − = −
f(-4) = -90 < 0 → satisfies
f(x) < 0Each factor's sign at x = -4: (x + 2) is negative; (x − 1) is negative; (x − 5) is negative. Combining them gives f(-4) = -90, which is indeed negative. Since the inequality asks for f(x) < 0, this x is in the solution set.