(x-5)^2(x+1)<0
1) Let's solve in ℝ: (x -5)2·(x + 1) < 0
We study the sign of each factor of the product
⇥1.1) Let's study the sign of (x -5)2 by widening the sign to ≤ and solving in ℝ: (x -5)2 ≤ 0
⇥Any even power is always non-negative in ℝ and equals 0 only when the base itself is 0
⇥1.2) Let's solve in ℝ: x -5 = 0
⇥ⓘ We add 5 to both sides
⇥Solution found: x = 5
⇥So the expression (x -5)2 is zero at x = 5 and positive otherwise
⇥The expression x + 1 is zero at x = -1, it is negative before and positive after
2) Let's draw up the sign table:
x-∞ -1 5 +∞
(x -5)2++0+
x + 1-0++
(x -5)2·(x + 1)-0+0+
▶Solution obtained: x < -1