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Understanding how to work with squared binomials is a fundamental skill in algebra. Squaring a binomial involves multiplying the binomial by itself. For example, \( (x-1)^2 \) translates to \( (x-1)(x-1) \). It’s like saying you need two of the same binomials and you are going to multiply them together.

When you start with this process, you should envision having two sets of parentheses, side by side, ready to be multiplied. This is crucial because it sets the stage for using the distributive property, which we will talk about next. The most common mistake students make when dealing with squared binomials is not correctly applying the distributive property to each term, leading to incorrect solutions.

The power of the distributive property lies in its ability to simplify the multiplication process, especially with binomials. When handling expressions like \( x\cdot(x-1) \) and \( -1\cdot(x-1) \) in our squared binomial \( (x-1)^2 \), it is vital to distribute each term of the first binomial across the terms of the second binomial.

Imagine each term in the first binomial stretching out an arm to distribute itself to each term in the second binomial. So for \( (x-1)(x-1) \), \( x \) is multiplied by \( x \) and then \( x \) by \( -1 \) and likewise for \( -1 \) in the first binomial. This action creates four products that we later combine. Without properly distributing, you cannot progress to the step of combining like terms to simplify your answer.

After distributing, combining like terms is like finding friends in a crowded room—you gather all the similar terms together to simplify the expression. In the case of \( x^2 - x - x + 1 \), after using the distributive property, we notice there are two \( -x \) terms. These are like terms because they share the same variable raised to the same power.

When we combine them, we add their coefficients (the numerical parts) together, which in this case results in \( -2x \). This simplification is crucial for arriving at the final, more manageable expression of \( x^2 - 2x + 1 \). Being alert and accurate during this step is essential, as combining like terms incorrectly can lead to an incorrect solution. Always keep an eye out for terms that look alike—they're your 'like terms' and they belong together.