# What are the zeros of the function g(x) = (x+5)(x-2)(x+1)

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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.**

g(x) =(x+5)(x-2)(x+1).

To find the zeros of the g(x).

We know that the zeros of a function g(x) are the set of values of x for which g(x) evaluates to zero.

If x1 is a zero g(x), then if we substitute x1 for x in g(x) , then g(x1) = 0.

Therefore if we can write g(x) in the form g(x) = (x-a)(x-b)(x-c), then g(a) = (a-a)(a-b)(a-c). Or g(a) = 0*(a-c)(a-c) = 0. That proves a zero of g(x).

Similarly if g(x) = (x-a)(x-b)(x-c), then g(b) = (b-a)(b-b)(b-c) = 0 , and g(c) = (c-a)(c-b)(c-c) = 0.

Therefore ifg(x) = (x-a)(x-b)(x-c), then x =a ,x = b and x= c are the zeros of g(x).

In this case g(x) = (x+5)(x+1)(x-2). So a = -5, b = -1 c = 2 are the zeros of g(x). The zeros are also the solutions of each factor of g(x) equated to zero: x+5 = 0 , x-2 = 0 , x-2 = 0

The zeroes of a function are values of x for which the values of the function is equal to zero.

Here we have: g(x) = (x+5)(x-2)(x+1)

Now if g(x)= 0

=>(x+5)(x-2)(x+1) = 0

=> x+5 = 0 or x -2 =0 or x +1 =0

=> x = -5 or x = 2 or x = -1

**Therefore the zeroes of the function are x = -5, x =2 and x = -1.**