1. **Problem statement:** (a) Find the highest common factor (HCF) of $5A$ and $2B$ given: $$A = 2^5 \times 5 \times 7^2$$ $$B = 2^3 \times 5^3 \times 7^4$$ (b) Calculate the value of $(AB)^2$ as a product of prime factors. 2. **Step (a) - Find HCF of $5A$ and $2B$:** - First express $5A$ and $2B$ in prime factor form: $$5A = 5 \times (2^5 \times 5 \times 7^2) = 2^5 \times 5^2 \times 7^2$$ $$2B = 2 \times (2^3 \times 5^3 \times 7^4) = 2^4 \times 5^3 \times 7^4$$ - The HCF is the product of the lowest powers of common prime factors: - For 2: $\min(5,4) = 4$ - For 5: $\min(2,3) = 2$ - For 7: $\min(2,4) = 2$ - Therefore: $$\text{HCF} = 2^4 \times 5^2 \times 7^2$$ 3. **Step (b) - Calculate $(AB)^2$:** - First find $AB$: $$AB = (2^5 \times 5 \times 7^2) \times (2^3 \times 5^3 \times 7^4) = 2^{5+3} \times 5^{1+3} \times 7^{2+4} = 2^8 \times 5^4 \times 7^6$$ - Now square $AB$: $$ (AB)^2 = (2^8)^2 \times (5^4)^2 \times (7^6)^2 = 2^{16} \times 5^8 \times 7^{12}$$ **Final answers:** - (a) $\boxed{2^4 \times 5^2 \times 7^2}$ - (b) $\boxed{2^{16} \times 5^8 \times 7^{12}}$