Answer :
To find the restrictions for the expression [tex]\(\frac{2x+6}{x^2+3x-18} \times \frac{x-4}{x^2-9}\)[/tex], we need to identify the values of [tex]\(x\)[/tex] that make the denominator of the entire expression equal to zero. These values are the ones that [tex]\(x\)[/tex] cannot be because they make the expression undefined.
### Step 1: Combine the Denominators
The denominators in the given expression are:
1. [tex]\(x^2 + 3x - 18\)[/tex]
2. [tex]\(x^2 - 9\)[/tex]
The overall denominator can be written as:
[tex]\[ (x^2 + 3x - 18)(x^2 - 9) \][/tex]
### Step 2: Find the Zeros of Each Denominator
To find the restrictions, let's solve for [tex]\(x\)[/tex] when the denominators are equal to zero.
#### Denominator 1: [tex]\(x^2 + 3x - 18\)[/tex]
We factorize this quadratic expression:
[tex]\[ x^2 + 3x - 18 = (x + 6)(x - 3) \][/tex]
Setting each factor to zero gives us two solutions:
[tex]\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
#### Denominator 2: [tex]\(x^2 - 9\)[/tex]
We recognize this as a difference of squares:
[tex]\[ x^2 - 9 = (x + 3)(x - 3) \][/tex]
Setting each factor to zero gives us two solutions:
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
### Step 3: Compile All Restrictions
By combining the solutions from both denominators, we find the restrictions are:
[tex]\[ x \neq -6, \quad x \neq -3, \quad x \neq 3 \][/tex]
Thus, the restrictions for the given expression are:
- [tex]\(x \neq -3\)[/tex]
- [tex]\(x \neq 3\)[/tex]
- [tex]\(x \neq -6\)[/tex]
### Step 4: Identify the Correct Options
Among the given options, the correct restrictions are included in the following options:
1. [tex]\(x \neq -3, 3\)[/tex]
2. [tex]\(x \neq 3\)[/tex]
3. [tex]\(x \neq -6\)[/tex]
Therefore, the correct selections are:
- [tex]\(x \neq -3, 3\)[/tex]
- [tex]\(x \neq 3\)[/tex]
- [tex]\(x \neq -6\)[/tex]
The incorrect restriction is:
- [tex]\(x \neq 4\)[/tex], as 4 does not make the denominator zero.
So the set of correct choices is:
[tex]\[ \text{Select } x \neq -3, \quad x \neq 3, \quad x \neq -6 \][/tex]
### Step 1: Combine the Denominators
The denominators in the given expression are:
1. [tex]\(x^2 + 3x - 18\)[/tex]
2. [tex]\(x^2 - 9\)[/tex]
The overall denominator can be written as:
[tex]\[ (x^2 + 3x - 18)(x^2 - 9) \][/tex]
### Step 2: Find the Zeros of Each Denominator
To find the restrictions, let's solve for [tex]\(x\)[/tex] when the denominators are equal to zero.
#### Denominator 1: [tex]\(x^2 + 3x - 18\)[/tex]
We factorize this quadratic expression:
[tex]\[ x^2 + 3x - 18 = (x + 6)(x - 3) \][/tex]
Setting each factor to zero gives us two solutions:
[tex]\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
#### Denominator 2: [tex]\(x^2 - 9\)[/tex]
We recognize this as a difference of squares:
[tex]\[ x^2 - 9 = (x + 3)(x - 3) \][/tex]
Setting each factor to zero gives us two solutions:
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
### Step 3: Compile All Restrictions
By combining the solutions from both denominators, we find the restrictions are:
[tex]\[ x \neq -6, \quad x \neq -3, \quad x \neq 3 \][/tex]
Thus, the restrictions for the given expression are:
- [tex]\(x \neq -3\)[/tex]
- [tex]\(x \neq 3\)[/tex]
- [tex]\(x \neq -6\)[/tex]
### Step 4: Identify the Correct Options
Among the given options, the correct restrictions are included in the following options:
1. [tex]\(x \neq -3, 3\)[/tex]
2. [tex]\(x \neq 3\)[/tex]
3. [tex]\(x \neq -6\)[/tex]
Therefore, the correct selections are:
- [tex]\(x \neq -3, 3\)[/tex]
- [tex]\(x \neq 3\)[/tex]
- [tex]\(x \neq -6\)[/tex]
The incorrect restriction is:
- [tex]\(x \neq 4\)[/tex], as 4 does not make the denominator zero.
So the set of correct choices is:
[tex]\[ \text{Select } x \neq -3, \quad x \neq 3, \quad x \neq -6 \][/tex]