This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions for Activity 1: 1.1 Write down all the possible factors for each of the following expressions. Step 1: List all factors for 3a^2b. The prime factorization of 3a^2b is 3 · a · a · b. The possible factors are combinations of these prime factors: 3a^2b = 1, 3, a, b, a^2, ab, a^2b, 3a, 3b, 3a^2, 3ab, 3a^2b Step 2: List all factors for 4ab^2. The prime factorization of 4ab^2 is 2 · 2 · a · b · b. The possible factors are combinations of these prime factors: 4ab^2 = 1, 2, 4, a, b, b^2, 2a, 2b, 2b^2, 4a, 4b, 4b^2, ab, ab^2, 2ab, 2ab^2, 4ab, 4ab^2 1.2 Determine the H.C.F. of the algebraic expression. Step 1: Find the H.C.F. of the numerical coefficients. The numerical coefficients are 3 and 4. Factors of 3: 1, 3 Factors of 4: 1, 2, 4 The H.C.F. of 3 and 4 is 1. Step 2: Find the H.C.F. of the variable a. The terms have a^2 and a^1. The lowest power of a is a^1. The H.C.F. of a^2 and a is a. Step 3: Find the H.C.F. of the variable b. The terms have b^1 and b^2. The lowest power of b is b^1. The H.C.F. of b and b^2 is b. Step 4: Combine the H.C.F. of the numerical and variable parts. The H.C.F. of the algebraic expression 3a^2b + 4ab^2 is the product of the H.C.F.s found in Steps 1, 2, and 3. H.C.F. = 1 × a × b = ab Last free one today — make it count tomorrow, or type /upgrade for unlimited.