Solve (x-4)² = (x+2)(x-1): Comparing Squared and Factored Forms

Let's examine the given equation:

$(x-4)^2=(x+2)(x-1)$First, let's simplify the equation, using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$and the expanded distributive law,

We'll start by opening the parentheses using the perfect square binomial formula mentioned and using the expanded distributive law and then we'll move terms and combine like terms:

$(x-4)^2=(x+2)(x-1) \\ \downarrow\\ x^2-2\cdot x\cdot4+4^2=x^2-x+2x-2\\ x^2-8x+16=x^2-x+2x-2\\ x^2-8x+16-x^2+x-2x+2=0\\ -9x+18=0$We got a first-degree equation, we'll solve it in the regular way by isolating the variable on one side:

$-9x+18=0 \\ -9x=-18\hspace{6pt}\text{/}:(-9)\\ \boxed{x=2}$

Let's summarize the equation solving steps:

$(x-4)^2=(x+2)(x-1) \\ \downarrow\\ x^2-8x+16=x^2-x+2x-2\\ -9x+18=0 \\ \downarrow\\ \boxed{x=2}$Therefore, the correct answer is answer C.