1) Factor the polynomial
x^2 + x - 6 = (x + 3) (x - 2)
So, can solve (x + 3)(x - 2) ≤ 0
The condition for the product of two factors be zero or negative is that the both are zero or have different signs (one positive and the other negative)
a) when x + 3 ≤ 0 and x - 2 ≥ 0
x + 3 ≤ 0 => x ≤ - 3
x - 2 ≥ 0 => x ≥ 2
Given that it is imposible that x be less or equal tha - 3 and greaer or equal than 2, this solution is empy.
b) when x + 3 ≥ 0 and x - 2 ≤ 0
x + 3 ≥ 0 => x ≥ - 3
x - 2 ≤ 0 => x ≤ 2
Solution: - 3 ≤ x ≤ 2 or [-3,2]
