Question

x^3 + 4x^2 - 4x - 16 ≥ 0

Answer 1

x^3 + 4x^2 - 4x - 16
x^2(x + 4) - 4(x + 4)
(x + 4)(x^2 - 4)
[(x + 4)(x + 2)(x - 2)\ge0

Answer 2

first write
\[(x+4) (x-2)(x+2) \geq 0\]

the zeros are at -4,-2, 2and then since this is a polynomial of degree 3 with positive leading coefficient it will be negative, then positive, then negative then positive

Answer 3

you want positive, so the answer is
\[[-4,-2]\cup [2,\infty)\] but of course you know where it it positive and where it is negative