To find the restricted values for the expression \(\frac{x^2-4}{4 x^2-12 x} \cdot \frac{2 x+4}{-x^2+5 x-6}\), we must determine the values of \(x\) that make any denominator equal to zero, as division by zero is undefined.
Let's consider the denominators in the given expression:
1. The first denominator is \(4x^2 - 12x\).
2. The second denominator is \(-x^2 + 5x - 6\).
We will find the values of \(x\) that make each denominator equal to zero.
### Step 1: Solve \(4x^2 - 12x = 0\)
First, factor this quadratic expression:
[tex]\[4x^2 - 12x = 4x(x - 3) = 0\][/tex]
Set each factor equal to zero:
[tex]\[4x = 0 \quad \Rightarrow \quad x = 0\][/tex]
[tex]\[x - 3 = 0 \quad \Rightarrow \quad x = 3\][/tex]
So, the values of \(x\) that make this denominator zero are \(x = 0\) and \(x = 3\).
### Step 2: Solve \(-x^2 + 5x - 6 = 0\)
We can rewrite and factor this quadratic expression:
[tex]\[ -x^2 + 5x - 6 = 0 \][/tex]
Multiply the equation by \(-1\) for easier factoring:
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
Factor the quadratic expression:
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
So, the values of \(x\) that make this denominator zero are \(x = 2\) and \(x = 3\).
### Combine Restricted Values
From both steps, the restricted values of \(x\) are \(x = 0\), \(x = 2\), and \(x = 3\).
### Final Answer
Combine the restricted values:
[tex]\[ \boxed{0, 2, 3} \][/tex]
Therefore, the correct answer is:
[tex]\[x \neq 0, x \neq 2, x \neq 3\][/tex]
Thus, the correct choice is:
B. [tex]\(x \neq 0, \neq 2, \neq 3\)[/tex]
