Let f(x) = 6x³ + 16x² + 7x - 2
When x = -2, f(-2) = 6(-2)³ + 16(-2)² + 7(-2) - 2
= -48 + 64 - 14 - 2
= 0
By the Factor Theorem, (x + 2) is a factor f(x).
Next, 6x³ + 16x² + 7x - 2 ≥ (x + 2)²
Factorise f(x), (you can either do long division or compare coefficients)
(x + 2)(6x² + 4x - 1) ≥ (x + 2)²
(x + 2)(6x² + 4x - 1) - (x + 2)² ≥ 0
(x + 2)(6x² + 4x - 1 - (x + 2)) ≥ 0
(x + 2)(6x² + 3x - 3) ≥ 0
(x + 2)(2x² + x - 1) ≥ 0
(x + 2)(2x - 1)(x + 1) ≥ 0
Test for the signs to the left and right of x = -2, x = ½ and x = -1,
x ≥ ½ or -2 ≤ x ≤ -1 for (x + 2)(2x - 1)(x + 1) to be ≥ 0
∴ Set of values is {x : x ∈ R , x ≥ ½ or -2 ≤ x ≤ -1}