Find the zeros of the polynomial function (show all work) f(x)=x^3 x^2-4x-4
( answer asap)

The zeros are -2, -1 and 2.

Step-by-step explanation:

I will assume that it is:

x^3 + x^2 - 4x - 4 = 0

Looks like x = -1  might be a root:

(-1)^3 + (-1)^2 - 4(-1) - 4

= - 1 + 1 + 4 - 4 = 0

So x = -1 is one zero.

As x = -1 is a zero then (x + 1) must be a factor of the expression so we divide:-

x + 1 ( x^3 + x^2 - 4x - 4  ( x^2 - 4  <  The quotient

         x^3 + x^2

                         -4x - 4

                          -4x - 4

So we have (x + 1)(x^2 - 4) = 0

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

Therefore  the zeroes are  -1,2 and -2,

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: