Find all of the zeros of the polynomial function and state the multiplicity of each.
f (x) = (x^2 – 16)^2
– 4 with multiplicity 2 and 4 with multiplicity 2
I think this is the ans for this equation. Anyone care to chime in? Thanks!
Yes, -4 and +4 each with multiplicity 2.
I think it is choice A, right?
To find the zeros of the polynomial function f(x), we need to solve the equation f(x) = 0. In this case, the polynomial function is f(x) = (x^2 - 16)^2.
First, let's factor the polynomial as much as possible. We have (x^2 - 16)^2, which can be further factored as [(x - 4)(x + 4)]^2.
Setting this equal to zero, we have [(x - 4)(x + 4)]^2 = 0. Since a squared term is zero if and only if its base is zero, we have (x - 4)(x + 4) = 0.
Now we can solve for x, considering the two factors separately:
1. (x - 4) = 0
Solving this equation, we get x = 4.
2. (x + 4) = 0
Solving this equation, we get x = -4.
Therefore, the zeros of the polynomial function f(x) = (x^2 - 16)^2 are x = -4 and x = 4.
Next, let's determine the multiplicity of each zero. The multiplicity of a zero is the number of times the factor (x - a) appears in the polynomial expression.
In this case, both zeros have a multiplicity of 2. This means that the factors (x - 4) and (x + 4) appear twice each in the polynomial expression (x^2 - 16)^2.
So, the zeros of the polynomial function f(x) = (x^2 - 16)^2 are -4 with multiplicity 2 and 4 with multiplicity 2.