16. Write the prime factorization of 120 . \( \begin{array}{lll}\text { F. } 2^{3} \cdot 3 \cdot 5 & \text { G. } 2^{3} \bullet 15 \\ \text { H. } 2^{2} \cdot 3^{2} \bullet 5 & \text { J. } 2 \cdot 3 \cdot 5\end{array} \)

To find the prime factorization of **120**, let's break it down step by step: 1. **Divide by 2 (the smallest prime number):** \[ 120 \div 2 = 60 \] 2. **Continue dividing by 2:** \[ 60 \div 2 = 30 \] 3. **Divide by 2 once more:** \[ 30 \div 2 = 15 \] 4. **Now, divide by the next prime number, which is 3:** \[ 15 \div 3 = 5 \] 5. **Finally, 5 is a prime number:** \[ 5 \div 5 = 1 \] Putting it all together, the prime factors of 120 are: \[ 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \cdot 3 \cdot 5 \] **Matching with the provided options:** - **F.** \( 2^{3} \cdot 3 \cdot 5 \) **→ Correct** - G. \( 2^{3} \bullet 15 \) (15 is not fully factored into primes) - H. \( 2^{2} \cdot 3^{2} \cdot 5 \) (This equals 180, not 120) - J. \( 2 \cdot 3 \cdot 5 \) (This equals 30, not 120) **Therefore, the correct prime factorization of 120 is option F:** \[ 2^{3} \cdot 3 \cdot 5 \]