The given inequality is expressed as
(x + 1)(x - 3)^2 ≥ 0
We would replace the inequality symbol with equal to. We have
(x + 1)(x - 3)^2 = 0
This can also be written as
(x + 1)(x - 3)(x - 3) = 0
This means that
x + 1 = 0 or x - 3 = 0 twice
x = - 1 or x = 3 twice
We would substitute values of x to the left and right of these roots into the inequality and check if they satisfy the inequality. Thus,
For x = - 2, we have
(- 2 + 1)(- 2 - 3)^2 ≥ 0
- 1)(-5)^2 ≥ 0
- 5 ≥ 0
We know that this is not true because - 5 is never greater than 0. Thus, values of x below - 1 does not satisfy the inequality
For x = 0, we have
(0 + 1)(0 - 3)^2 ≥ 0
1)(-3)^2 ≥ 0
3 ≥ 0
This is true
For x = 4, we have
(4 + 1)(4 - 3)^2 ≥ 0
5)(1)^2 ≥ 0
5 ≥ 0
This is true
This means that all values of x to the right of - 1 satisfy the inequality. This means that the solutions are between - 1 and infinity. The correct option is the last one. The bracket is used because - 1 is inclusive
[- 1, infinity)