To determine the zeros of the function [tex]\(F(x) = \frac{x(x-2)}{(x+3)(x-5)}\)[/tex], we need to follow these steps:
1. Identify the zeros of the numerator:
The zeros of the numerator are the values of [tex]\(x\)[/tex] that make the numerator zero. In the function [tex]\(F(x)\)[/tex], the numerator is [tex]\(x(x-2)\)[/tex].
Set the numerator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x(x-2) = 0
\][/tex]
This equation can be solved by setting each factor to zero:
[tex]\[
x = 0 \quad \text{or} \quad x-2 = 0 \implies x = 2
\][/tex]
Therefore, the zeros of the numerator are [tex]\(x = 0\)[/tex] and [tex]\(x = 2\)[/tex].
2. Identify the zeros of the denominator:
The denominator must not be zero for the function to be defined. The denominator of [tex]\(F(x)\)[/tex] is [tex]\((x+3)(x-5)\)[/tex].
Set the denominator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
(x+3)(x-5) = 0
\][/tex]
This equation can be solved by setting each factor to zero:
[tex]\[
x+3 = 0 \implies x = -3 \quad \text{or} \quad x-5 = 0 \implies x = 5
\][/tex]
Therefore, the zeros of the denominator (which we need to exclude) are [tex]\(x = -3\)[/tex] and [tex]\(x = 5\)[/tex].
3. Determine the valid zeros of the function:
Valid zeros of the function are the zeros of the numerator that are not zeros of the denominator. From step 1, the potential zeros are [tex]\(x = 0\)[/tex] and [tex]\(x = 2\)[/tex]. From step 2, we exclude [tex]\(x = -3\)[/tex] and [tex]\(x = 5\)[/tex].
Since [tex]\(0\)[/tex] and [tex]\(2\)[/tex] are not excluded by the denominator, they remain as valid zeros.
4. Conclusion:
The zeros of the function [tex]\(F(x) = \frac{x(x-2)}{(x+3)(x-5)}\)[/tex] are:
- [tex]\(x = 0\)[/tex]
- [tex]\(x = 2\)[/tex]
Therefore, the correct answers to the question are:
- A. [tex]\(2\)[/tex]
- B. [tex]\(0\)[/tex]
So the zeros of the function are [tex]\(0\)[/tex] and [tex]\(2\)[/tex].
