Factor the trinomial. \( 14 a^{3}-33 a^{2}+18 a \) Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The answer is B. The trinomial is not factorable.

To factor the trinomial \( 14 a^{3}-33 a^{2}+18 a \), first, we can factor out the greatest common factor, which is \( a \): \( a(14 a^{2} - 33 a + 18) \). Now, we need to factor the quadratic \( 14 a^{2} - 33 a + 18 \). We can look for two numbers that multiply to \( 14 \times 18 = 252 \) and add up to \( -33 \). After testing combinations, we find that \( -9 \) and \( -28 \) work: Thus, rewrite the quadratic as: \( 14 a^{2} - 9 a - 24 a + 18 \). Then, group and factor by grouping: \( (14 a^{2} - 9 a) - (24 a - 18) \) which becomes \( 7a(2a - 9) - 2(2a - 9) \). Now we can factor it as: \( (2a - 9)(7a - 2) \). Putting it all together, the fully factored form of the original trinomial is: \( a(2a - 9)(7a - 2) \). So, the answer is: A. The answer is \( a(2a - 9)(7a - 2) \).