`{:("Column A",x != 3 and x != 6,"Column B"),((2x^2 - 72)/(x - 6),,(2x^2 - 18)/(x - 3)):}`

Options

A. If column A is larger
B. If column B is larger
C. If the columns are equal
D. If there is not enough information to decide

Answer

To solve the problem, we need to simplify the expressions in Column A and Column B and then compare their values.

### Step 1: Simplify Column A
We start with the expression in Column A:
\[ \text{Column A} = \frac{2x^2 - 72}{x - 6} \]
First, we can factor out the numerator:
\[ 2x^2 - 72 = 2(x^2 - 36) = 2(x^2 - 6^2) \]
Now we can use the difference of squares:
\[ x^2 - 6^2 = (x + 6)(x - 6) \]
Thus, we can rewrite Column A:
\[ \text{Column A} = \frac{2(x + 6)(x - 6)}{x - 6} \]
Since \(x \neq 6\), we can cancel \(x - 6\):
\[ \text{Column A} = 2(x + 6) \]

### Step 2: Simplify Column B
Now we simplify the expression in Column B:
\[ \text{Column B} = \frac{2x^2 - 18}{x - 3} \]
We can factor out the numerator:
\[ 2x^2 - 18 = 2(x^2 - 9) = 2(x^2 - 3^2) \]
Using the difference of squares again:
\[ x^2 - 3^2 = (x + 3)(x - 3) \]
Thus, we can rewrite Column B:
\[ \text{Column B} = \frac{2(x + 3)(x - 3)}{x - 3} \]
Since \(x \neq 3\), we can cancel \(x - 3\):
\[ \text{Column B} = 2(x + 3) \]

### Step 3: Compare Column A and Column B
Now we have:
\[ \text{Column A} = 2(x + 6) \]
\[ \text{Column B} = 2(x + 3) \]
To compare these, we can look at the difference:
\[ \text{Column A} - \text{Column B} = 2(x + 6) - 2(x + 3) = 2x + 12 - 2x - 6 = 6 \]
Since \(6 > 0\), it follows that:
\[ \text{Column A} > \text{Column B} \]

### Conclusion
Thus, Column A is greater than Column B for all \(x\) such that \(x \neq 3\) and \(x \neq 6\).

### Final Answer
The answer is that Column A is greater than Column B.

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