4. Find the prime factorisation of each of the following numbers, expressing your answer in index notation \( \begin{array}{ll}\text { (a) } 24 & \text { (b) } 95 \\ \text { (c) } 225 & \text { (d) } 442\end{array} \)

To find the prime factorization of each number and express the answer in index notation, we will break down each number into its prime factors. (a) **24**: - Start by dividing by the smallest prime number, which is 2: - \( 24 \div 2 = 12 \) - \( 12 \div 2 = 6 \) - \( 6 \div 2 = 3 \) - \( 3 \div 3 = 1 \) - The prime factors are \( 2, 2, 2, \) and \( 3 \). - In index notation, this is: \[ 24 = 2^3 \times 3^1 \] (b) **95**: - Start by dividing by the smallest prime number: - \( 95 \div 5 = 19 \) - 19 is a prime number. - The prime factors are \( 5 \) and \( 19 \). - In index notation, this is: \[ 95 = 5^1 \times 19^1 \] (c) **225**: - Start by dividing by the smallest prime number: - \( 225 \div 3 = 75 \) - \( 75 \div 3 = 25 \) - \( 25 \div 5 = 5 \) - \( 5 \div 5 = 1 \) - The prime factors are \( 3, 3, 5, \) and \( 5 \). - In index notation, this is: \[ 225 = 3^2 \times 5^2 \] (d) **442**: - Start by dividing by the smallest prime number: - \( 442 \div 2 = 221 \) - \( 221 \div 13 = 17 \) - 17 is a prime number. - The prime factors are \( 2, 13, \) and \( 17 \). - In index notation, this is: \[ 442 = 2^1 \times 13^1 \times 17^1 \] In summary, the prime factorizations in index notation are: - (a) \( 24 = 2^3 \times 3^1 \) - (b) \( 95 = 5^1 \times 19^1 \) - (c) \( 225 = 3^2 \times 5^2 \) - (d) \( 442 = 2^1 \times 13^1 \times 17^1 \)