4. Find the prime factorisation of each of the following numbers, expressing your answer in index notation. \( \begin{array}{ll}\text { (a) } 24 \\ 2 \sqrt{27} \\ 2 \sqrt{\frac{1}{2}} & \text { (b) } 95 \\ 3 \sqrt{225} & \text { (d) } 442 \\ 3 \longdiv { 7 5 } & 2 \sqrt{44} \\ 5 \sqrt{25} & \text { (f) } 1200\end{array} \)

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 1. Start with 24: - \( 24 = 2 \times 12 \) - \( 12 = 2 \times 6 \) - \( 6 = 2 \times 3 \) So, \( 24 = 2^3 \times 3^1 \). ### \( 2 \sqrt{27} \) 1. Calculate \( \sqrt{27} \): - \( 27 = 3^3 \) - Therefore, \( \sqrt{27} = 3^{3/2} = 3^{1.5} = 3^{1} \times 3^{0.5} \). 2. Now, combine with 2: - \( 2 \sqrt{27} = 2^1 \times 3^{3/2} \). ### \( 2 \sqrt{\frac{1}{2}} \) 1. Calculate \( \sqrt{\frac{1}{2}} \): - \( \frac{1}{2} = 2^{-1} \) - Therefore, \( \sqrt{\frac{1}{2}} = \sqrt{2^{-1}} = 2^{-1/2} \). 2. Now, combine with 2: - \( 2 \sqrt{\frac{1}{2}} = 2^1 \times 2^{-1/2} = 2^{1 - 1/2} = 2^{1/2} \). ### (b) 95 1. Start with 95: - \( 95 = 5 \times 19 \) (both 5 and 19 are prime). So, \( 95 = 5^1 \times 19^1 \). ### \( 3 \sqrt{225} \) 1. Calculate \( \sqrt{225} \): - \( 225 = 15^2 = (3 \times 5)^2 = 3^2 \times 5^2 \). - Therefore, \( \sqrt{225} = 15 = 3^1 \times 5^1 \). 2. Now, combine with 3: - \( 3 \sqrt{225} = 3^1 \times (3^1 \times 5^1) = 3^{1+1} \times 5^1 = 3^2 \times 5^1 \). ### (d) 442 1. Start with 442: - \( 442 = 2 \times 221 \) - \( 221 = 13 \times 17 \) (both 13 and 17 are prime). So, \( 442 = 2^1 \times 13^1 \times 17^1 \). ### \( 3 \longdiv { 7 5 } \) 1. Calculate \( 3 \div 75 \): - \( 75 = 3 \times 25 = 3 \times 5^2 \). - Therefore, \( 3 \div 75 = \frac{3^1}{3^1 \times 5^2} = 3^{1-1} \times 5^{-2} = 5^{-2} \). ### \( 2 \sqrt{44} \) 1. Calculate \( \sqrt{44} \): - \( 44 = 4 \times 11 = 2^2 \times 11 \). - Therefore, \( \sqrt{44} = \sqrt{2^2 \times 11} = 2^1 \times 11^{1/2} \). 2. Now, combine with 2: - \( 2 \sqrt{44} = 2^1 \times (2^1 \times 11^{1/2}) = 2^{1+1} \times 11^{1/2} = 2^2 \times 11^{1/2} \). ### (f) 1200 1. Start with 1200: - \( 1200 = 12 \times 100 = (3 \times 2^2) \times (10^2) = (3 \times 2^2) \times (2 \times 5)^2 \). - Therefore, \( 1200 = 3^1 \times 2^2 \times (2^2 \times 5^2) = 3^