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 factorisations: 2.1 4x^2 - 9c^6 Step 1: Recognize this as a difference of squares, a^2 - b^2 = (a - b)(a + b). Here, a = sqrt(4x^2) = 2x and b = sqrt(9c^6) = 3c^3. 4x^2 - 9c^6 = (2x)^2 - (3c^3)^2 = (2x - 3c^3)(2x + 3c^3) The factorised expression is (2x - 3c^3)(2x + 3c^3). 2.2 -7a^2b + 28b Step 1: Find the greatest common factor (GCF) of the terms. The GCF of -7 and 28 is 7. The GCF of a^2b and b is b. So, the common factor is 7b. We can factor out -7b to make the leading term positive. -7a^2b + 28b = -7b(a^2) + (-7b)(-4) = -7b(a^2 - 4) Step 2: Recognize that (a^2 - 4) is a difference of squares, a^2 - b^2 = (a - b)(a + b). Here, a = a and b = sqrt(4) = 2. = -7b(a - 2)(a + 2) The factorised expression is -7b(a - 2)(a + 2). 2.3 n^4 - 1 Step 1: Recognize this as a difference of squares, a^2 - b^2 = (a - b)(a + b). Here, a = sqrt(n^4) = n^2 and b = sqrt(1) = 1. n^4 - 1 = (n^2)^2 - (1)^2 = (n^2 - 1)(n^2 + 1) Step 2: Notice that (n^2 - 1) is another difference of squares. Here, a = sqrt(n^2) = n and b = sqrt(1) = 1. = (n - 1)(n + 1)(n^2 + 1) The factorised expression is (n - 1)(n + 1)(n^2 + 1). 2.4 (2x - 1)^2 - 49 Step 1: Recognize this as a difference of squares, a^2 - b^2 = (a - b)(a + b). Here, a = (2x - 1) and b = sqrt(49) = 7. (2x - 1)^2 - 49 = [(2x - 1) - 7][(2x - 1) + 7] Step 2: Simplify the terms inside the brackets. = (2x - 1 - 7)(2x - 1 + 7) = (2x - 8)(2x + 6) Step 3: Factor out any common factors from each binomial. From (2x - 8), the GCF is 2. So, 2(x - 4). From (2x + 6), the GCF is 2. So, 2(x + 3). = 2(x - 4) · 2(x + 3) = 4(x - 4)(x + 3) The factorised expression is 4(x - 4)(x + 3). 3 done, 2 left today. You're making progress.