You need to find the three solutions to the given equation, such that:

`(x+5)(x-2)(x+1) = 0 => {(x + 5 = 0),(x - 2 = 0),(x + 1 = 0):}`

`{(x = -5),(x = 2),(x = -1):}`

**Hence, evaluating the values of x for the equation `(x+5)(x-2)(x+1) = 0` holds, yields **`x = -5 ,x = -1, x = 2.`

The zeroes of the function are all the roots of the equation (x+5)(x-2)(x+1) = 0, putting g(x) = 0.

Since the expression of the function is a product, and, according to the rule, a product is zeri if at least one factor is zero, we'll set each factor as zero.

x + 5 = 0

We'll subtract 5 both sides:

x = -5

The first zero of g(x) is x = 5.

x - 2 = 0

We'll add 2 both sides:

x = 2

The second zero of g(x) is x = 2.

x+1 = 0

We'll subtract 1 both sides:

x = -1

The third zero of g(x) is x = -1.

Since the polynomial g(x) is of 3rd order, we cannot have more than 3 zeroes.