To determine the zeros of the function [tex]\( F(x) = \frac{(x+3)(x-1)}{(x-2)(x+2)} \)[/tex], we need to set the numerator equal to zero while ensuring that the denominator does not become zero. The numerator of the function is [tex]\((x + 3)(x - 1)\)[/tex] and the denominator is [tex]\((x - 2)(x + 2)\)[/tex].
### Step-by-Step Solution:
1. Find the zeros of the numerator:
We set the numerator [tex]\((x + 3)(x - 1)\)[/tex] equal to zero:
[tex]\[
(x + 3)(x - 1) = 0
\][/tex]
Solving for [tex]\(x\)[/tex], we get:
[tex]\[
x + 3 = 0 \quad \text{or} \quad x - 1 = 0
\][/tex]
[tex]\[
x = -3 \quad \text{or} \quad x = 1
\][/tex]
Therefore, the zeros are [tex]\(x = -3\)[/tex] and [tex]\(x = 1\)[/tex].
2. Check that the zeros do not make the denominator zero:
The denominator of the function is [tex]\((x - 2)(x + 2)\)[/tex]. We must ensure that neither [tex]\(x = -3\)[/tex] nor [tex]\(x = 1\)[/tex] makes the denominator zero:
- Substituting [tex]\(x = -3\)[/tex]:
[tex]\[
(x - 2)(x + 2) = (-3 - 2)(-3 + 2) = (-5)(-1) = 5 \neq 0
\][/tex]
- Substituting [tex]\(x = 1\)[/tex]:
[tex]\[
(x - 2)(x + 2) = (1 - 2)(1 + 2) = (-1)(3) = -3 \neq 0
\][/tex]
Both [tex]\(x = -3\)[/tex] and [tex]\(x = 1\)[/tex] do not make the denominator zero, so these values are valid zeros for the function.
3. Check the given options against the zeros:
The given options are: -5, -3, 3, 2, 1, -2. We need to check which of these are zeros of the function:
- Option [tex]\(-5\)[/tex]: Not a zero.
- Option [tex]\(-3\)[/tex]: Is a zero.
- Option [tex]\(3\)[/tex]: Not a zero.
- Option [tex]\(2\)[/tex]: Not a zero (moreover, it makes the denominator zero).
- Option [tex]\(1\)[/tex]: Is a zero.
- Option [tex]\(-2\)[/tex]: Not a zero (moreover, it makes the denominator zero).
Thus, the zeros of the function [tex]\( F(x) = \frac{(x+3)(x-1)}{(x-2)(x+2)} \)[/tex] from the given options are:
- B. -3
- E. 1
