Find the zeros of the function. State the multiplicity of multiple zeros.
y=6x^3-6x
To find the zeros of the function, we set y equal to zero and solve for x:
0 = 6x^3 - 6x
Next, we can factor out a common factor of 6x from both terms:
0 = 6x(x^2 - 1)
Now, we have two factors: 6x and (x^2 - 1).
Setting each factor equal to zero, we have:
6x = 0 --> x = 0
x^2 - 1 = 0
Using the difference of squares formula, we have:
(x - 1)(x + 1) = 0
Setting each factor equal to zero, we have:
x - 1 = 0 --> x = 1
x + 1 = 0 --> x = -1
Therefore, the zeros of the function are: x = 0, x = 1, and x = -1.
The multiplicity of each zero is 1, meaning each zero appears only once in the factored form of the function.
To find the zeros of the function y = 6x^3 - 6x, we set the function equal to zero and solve for x:
6x^3 - 6x = 0
Factoring out the common factor of 6x, we have:
6x(x^2 - 1) = 0
Now we can set each factor equal to zero and solve for x:
First factor:
6x = 0
x = 0
Second factor:
x^2 - 1 = 0
(x - 1)(x + 1) = 0
Setting each factor equal to zero gives us two more solutions:
x - 1 = 0 --> x = 1
x + 1 = 0 --> x = -1
So, the zeros of the function are x = 0, x = 1, and x = -1.
Now, let's determine the multiplicity of each zero.
For the zero x = 0, since it corresponds to a linear factor (6x), its multiplicity is 1.
For the zeros x = 1 and x = -1, they correspond to quadratic factors (x^2 - 1). Since the quadratic factors can be factored further, let's factor them to determine the multiplicity:
(x - 1)(x + 1) = 0
(x - 1)(x + 1) = 0
This factors completely, and we can see that each quadratic factor has multiplicity 1.
Therefore, the multiplicity of the zeros x = 0, x = 1, and x = -1 is 1 for each zero.