To determine the restrictions for the given expression [tex]\(\frac{6x+18}{x^2-4x+4} \times \frac{x+4}{x^2-4}\)[/tex], we need to examine the denominators and find the values of [tex]\(x\)[/tex] that would make the denominators zero, as these are the values for which the expression is undefined.
The expression has two denominators:
1. [tex]\(x^2 - 4x + 4\)[/tex]
2. [tex]\(x^2 - 4\)[/tex]
We will examine each denominator separately.
### Denominator 1: [tex]\(x^2 - 4x + 4\)[/tex]
This is a quadratic equation. We can factor it to find the roots:
[tex]\[x^2 - 4x + 4 = (x - 2)^2\][/tex]
Setting the expression equal to zero:
[tex]\[(x - 2)^2 = 0\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[x - 2 = 0\][/tex]
[tex]\[x = 2\][/tex]
So, [tex]\(x = 2\)[/tex] would make the first denominator zero. Thus, one restriction is [tex]\(x \neq 2\)[/tex].
### Denominator 2: [tex]\(x^2 - 4\)[/tex]
This is also a quadratic equation. We can factor it:
[tex]\[x^2 - 4 = (x - 2)(x + 2)\][/tex]
Setting the expression equal to zero:
[tex]\[(x - 2)(x + 2) = 0\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[x - 2 = 0 \quad \text{or} \quad x + 2 = 0\][/tex]
[tex]\[x = 2 \quad \text{or} \quad x = -2\][/tex]
So, [tex]\(x = 2\)[/tex] or [tex]\(x = -2\)[/tex] would make the second denominator zero. Thus, additional restrictions are [tex]\(x \neq 2\)[/tex] and [tex]\(x \neq -2\)[/tex].
### Summary of Restrictions
Combining the restrictions from both denominators, we have:
- [tex]\(x \neq 2\)[/tex]
- [tex]\(x \neq -2\)[/tex]
### Analysis of Given Options
The provided options are:
- [tex]\(x \neq -3\)[/tex]
- [tex]\(x \neq 2\)[/tex]
- [tex]\(x \neq -4\)[/tex]
- [tex]\(x \neq -2\)[/tex]
Based on our determined restrictions, [tex]\(x \neq 2\)[/tex] and [tex]\(x \neq -2\)[/tex] are correct.
Therefore, the restrictions for the expression are:
[tex]\[x \neq 2 \text{ and } x \neq -2\][/tex]
The correct options are:
- [tex]\(x \neq 2\)[/tex]
- [tex]\(x \neq -2\)[/tex]
