Determine the zeros of the function f(x) = (x - 2)(x + 2)(x+1)^2

x=2, x=-2, and x=-1

Step-by-step explanation:

Function: f(x)=(x-2)(x+2)(x+1)²

We can solve each one of the zeros by making the expressions (the ones in the parentheseis) equal zero.

So we have three equations:

1.(x-2)=0

2.(x+2)=0

3.(x+1)²=0

Solve.

1.x-2=0

x=2

2. x+2=0

x=-2

3. (x+1)²=(x+1)(x+1)=0

(x+1)=0

x=-1

(x+1)=0

x=-1

Since those two were the same we can just say x=-1 once.

Here are the three zeros:

x=2, x=-2, and x=-1

hope this helps!

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation:

step-by-step explanation: yes