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's the solution for Question 2.1: 2.1.1 Fully factorise each expression Step 1: Factorize the first expression. a^2 - 6ab + 9b^2 This is a perfect square trinomial of the form (x-y)^2 = x^2 - 2xy + y^2. Here, x=a and y=3b. a^2 - 6ab + 9b^2 = (a - 3b)^2 Step 2: Factorize the second expression. a^2 - 9b^2 This is a difference of squares of the form x^2 - y^2 = (x-y)(x+y). Here, x=a and y=3b. a^2 - 9b^2 = (a - 3b)(a + 3b) Step 3: Factorize the third expression. a^2 - 3ab Factor out the common term a. a^2 - 3ab = a(a - 3b) Step 4: Factorize the fourth expression. 3a^2 - 6ab - 9b^2 First, factor out the common numerical factor, which is 3. 3a^2 - 6ab - 9b^2 = 3(a^2 - 2ab - 3b^2) Now, factor the quadratic expression a^2 - 2ab - 3b^2. We look for two terms that multiply to -3b^2 and add to -2b. These are -3b and b. a^2 - 2ab - 3b^2 = (a - 3b)(a + b) So, the fully factorized expression is: 3a^2 - 6ab - 9b^2 = 3(a - 3b)(a + b) 2.1.2 Determine the LCM of all four expressions. Step 5: List all unique factors from the factorized expressions and identify their highest powers. The factorized expressions are: 1. (a - 3b)^2 2. (a - 3b)(a + 3b) 3. a(a - 3b) 4. 3(a - 3b)(a + b) The unique factors are: • Numerical factor: 3 (from expression 4) • Algebraic factors: a, (a - 3b), (a + 3b), (a + b) The highest powers for each unique factor are: • 3^1 • a^1 • (a - 3b)^2 (from expression 1) • (a + 3b)^1 (from expression 2) • (a + b)^1 (from expression 4) Step 6: Multiply all unique factors raised to their highest powers to find the LCM. LCM = 3 · a · (a - 3b)^2 · (a + 3b) · (a + b) LCM = 3a(a - 3b)^2(a + 3b)(a + b) Last free one today — make it count tomorrow, or type /upgrade for unlimited.