Question

Determine the prime factorizations of each number. a. 12 b. 75 c. 42 d. 96

Short answer

Answer: The prime factorizations are as follows: - 12: \(2^2\times 3^1\) - 75: \(3^1\times 5^2\) - 42: \(2^1\times 3^1\times 7^1\) - 96: \(2^5\times 3^1\)

Step-by-step solution

a. Prime factorization of 12

1. Start with the smallest prime number, 2, and divide 12 by 2: 12 ÷ 2 = 6. 2. Since 6 is not prime, continue dividing by the smallest prime factors: 6 ÷ 2 = 3 3. We now have all prime factors: 2, 2, and 3. So, the prime factorization of 12 is \(2^2\times 3^1\).

b. Prime factorization of 75

1. Start with the smallest prime number, 2, but 75 is an odd number and is not divisible by 2. 2. Move to next smallest prime number, 3, still 75 is not divisible by 3. 3. Move to next smallest prime number, 5, and divide 75 by 5: 75 ÷ 5 = 15. 4. Since 15 is not prime, continue dividing by smallest prime factors: 15 ÷ 3 = 5. 5. We now have all prime factors: 3, 5, and 5. So, the prime factorization of 75 is \(3^1\times 5^2\).

c. Prime factorization of 42

1. Start with the smallest prime number, 2, and divide 42 by 2: 42 ÷ 2 = 21. 2. Since 21 is not prime, move to next smallest prime number, 3, and divide 21 by 3: 21 ÷ 3 = 7. 3. We now have all prime factors: 2, 3, and 7. So, the prime factorization of 42 is \(2^1\times 3^1\times 7^1\).

d. Prime factorization of 96

1. Start with the smallest prime number, 2, and divide 96 by 2: 96 ÷ 2 = 48. 2. Since 48 is not prime, continue dividing by the smallest prime factors: 48 ÷ 2 = 24. 3. Continue dividing by smallest prime factors: 24 ÷ 2 = 12. 4. Continue dividing by smallest prime factors: 12 ÷ 2 = 6. 5. Continue dividing by smallest prime factors: 6 ÷ 2 = 3. 6. We now have all prime factors: 2, 2, 2, 2, 2, and 3. So, the prime factorization of 96 is \(2^5\times 3^1\).