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
f(x) = (x2-16)2
= ((x+4)(x-4))2
= (x+4)2(x-4)2
Can you take it from here?
Yes this is what I got, is it correct?
C. 4 with multiplicity 2 ?
No, it is not the case. There are four roots for a quartic equation, so one single root with multiplicity of 2 does not suffice.
When you have the factor (x+4)2, that implies x=-4 with multiplicity of 2.
If you repeat the process with the factor (x-4)2, you will find the answer you need.
Wow I am confused now. So does the answer include the i? which is
B. – 4i with multiplicity 2 and 4i with multiplicity 2
OR
A. – 4 with multiplicity 2 and 4 with multiplicity 2
I am goin to say A but I could be wrong.
A is correct. The roots are real, so there is no i involved.
There are two distinct roots, ±4 each with multiplicity of 2. So A is the answer.
To find the zeros of the polynomial function f(x) = (x^2 - 16)^2, we need to set the function equal to zero and solve for x.
First, let's expand the function using the property of a difference of squares: (a^2 - b^2) = (a - b)(a + b).
f(x) = (x - 4)(x + 4)^2
Now, we can set f(x) equal to zero:
(x - 4)(x + 4)^2 = 0
To find the zeros, we set each factor equal to zero and solve for x:
1. (x - 4) = 0
x = 4
2. (x + 4)^2 = 0
Taking the square root of both sides, we get:
x + 4 = 0
x = -4
So, the zeros of the function f(x) = (x^2 - 16)^2 are x = 4 and x = -4.
Now let's determine the multiplicity of each zero.
In this case, the multiplicity is determined by the power to which each factor is raised. For example, if a factor is raised to the power of 1, it has a multiplicity of 1; if it is raised to the power of 2, it has a multiplicity of 2, and so on.
From the expanded form of f(x), we can see that (x - 4) has multiplicity 1 and (x + 4)^2 has multiplicity 2.
Therefore, the answer is A. -4 with multiplicity 2 and 4 with multiplicity 2