1. **State the problem:** Find the prime factorization of the numbers 432, 740, 717, and 123. 2. **Recall the prime factorization method:** To factor a number into primes, divide it by the smallest prime numbers (2, 3, 5, 7, 11, ...) repeatedly until the quotient is 1. 3. **Factor 432:** - Divide by 2: $$432 \div 2 = 216$$ - Divide by 2: $$216 \div 2 = 108$$ - Divide by 2: $$108 \div 2 = 54$$ - Divide by 2: $$54 \div 2 = 27$$ - Divide by 3: $$27 \div 3 = 9$$ - Divide by 3: $$9 \div 3 = 3$$ - Divide by 3: $$3 \div 3 = 1$$ So, $$432 = 2^4 \times 3^3$$ 4. **Factor 740:** - Divide by 2: $$740 \div 2 = 370$$ - Divide by 2: $$370 \div 2 = 185$$ - Divide by 5: $$185 \div 5 = 37$$ - 37 is prime. So, $$740 = 2^2 \times 5 \times 37$$ 5. **Factor 717:** - Divide by 3: $$717 \div 3 = 239$$ - 239 is prime. So, $$717 = 3 \times 239$$ 6. **Factor 123:** - Divide by 3: $$123 \div 3 = 41$$ - 41 is prime. So, $$123 = 3 \times 41$$ **Final answers:** - (a) $$432 = 2^4 \times 3^3$$ - (b) $$740 = 2^2 \times 5 \times 37$$ - (c) $$717 = 3 \times 239$$ - (d) $$123 = 3 \times 41$$