A. \( 3 c^{2}-147=\square \) B. The binomial is not factorable.

Let's analyze the expression given in part A: A. \( 3c^{2} - 147 = \square \) To simplify this expression, we can factor out the common factor from the terms on the left side: 1. Notice that both terms \(3c^2\) and \(147\) have a common factor of \(3\). 2. Factor out \(3\): \[ 3(c^2 - 49) = \square \] 3. Now, we can recognize that \(c^2 - 49\) is a difference of squares, which can be factored further: \[ c^2 - 49 = (c - 7)(c + 7) \] 4. Therefore, we can rewrite the expression as: \[ 3(c - 7)(c + 7) = \square \] So, the expression \(3c^{2} - 147\) can be factored as \(3(c - 7)(c + 7)\). B. The statement "The binomial is not factorable" is not applicable here since we have shown that the expression \(3c^{2} - 147\) can indeed be factored. In conclusion: - A: \(3(c - 7)(c + 7)\) - B: The binomial \(3c^{2} - 147\) is factorable.