Factorise completely \( 2.2 .1 \quad\left[9-4(x-1)^{2}\right] \)

To factorise the expression \( 2.2.1 \quad\left[9-4(x-1)^{2}\right] \), we start by simplifying what’s inside the brackets. First, notice that \( 9 \) can be considered as \( 3^2 \), and \( 4(x-1)^{2} \) is a perfect square, so we can rewrite the expression as: \[ 9 - 4(x-1)^{2} = 3^2 - (2(x-1))^2 \] This resembles the difference of squares, which can be factorised using the formula \( a^2 - b^2 = (a - b)(a + b) \). In this case, \( a = 3 \) and \( b = 2(x-1) \): \[ = (3 - 2(x-1))(3 + 2(x-1)) \] Now substituting back and factoring the entire expression gives us: \[ 2.2.1 \quad [(3 - 2(x-1))(3 + 2(x-1))] \] Now, simplifying further, we get: \[ = 2.2.1 \quad [(3 - 2x + 2)(3 + 2x - 2)] \] \[ = 2.2.1 \quad [(5 - 2x)(1 + 2x)] \] This results in a completely factored form, showcasing the creative and logical aspects of algebraic manipulation!