Find all of the zeros of the polynomial function and state the multiplicity of each.

f (x) = (x^2 – 16)^2

A. – 4 with multiplicity 2 and 4 with multiplicity 2

B. – 4i with multiplicity 2 and 4i with multiplicity 2

C. 4 with multiplicity 2

D. 4 with multiplicity 4

I think the answer is A, any thoughts?

look at it this way...

f(x) = (x^2 – 16)^2
= (x^2 - 16)(x^# - 16)
= (x+4)(x-4)(x+4)(x-4)

the answer should be obvious, trust yourself !

I wasn't sure but it is A then?

yes,

how about that confidence thing ?

Not too good with math, actually i am no good at it.

To find the zeros of a polynomial function, you need to solve the equation f(x) = 0. In this case, the polynomial function is f(x) = (x^2 – 16)^2.

Let's first simplify the equation:

(x^2 – 16)^2 = 0

Now take the square root of both sides:

sqrt((x^2 – 16)^2) = sqrt(0)

|x^2 – 16| = 0

Since the square root of zero is zero, we can disregard the square root sign on the right side.

Now we can split the equation into two separate cases:

Case 1: x^2 – 16 = 0
Solving this equation gives us:
x^2 = 16
x = ± 4

So, in this case, we have two zeros: x = 4 and x = -4.

Case 2: -(x^2 – 16) = 0
Solving this equation gives us:
x^2 = 16
x = ± 4

So, in this case, we again have two zeros: x = 4 and x = -4.

Combining the results from both cases, we have the zeros x = 4 and x = -4, each with multiplicity 2.

Therefore, the correct answer is option A: – 4 with multiplicity 2 and 4 with multiplicity 2.