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

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