Factor the following trinomial. $$ \begin{array}{c}x^{2}-4 x+4 \ ([?] x-\square)^{2}\end{array} $$

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To factor the trinomial $$ x^{2} - 4x + 4 $$, we can look for a perfect square trinomial. A perfect square trinomial has the form $$(ax + b)^2 = a^2x^2 + 2abx + b^2$$.

Let's compare this with the given trinomial:

1. **Identify $$a$$:**  
   The coefficient of $$ x^2 $$ is 1, so $$ a = 1 $$.

2. **Identify $$b$$:**  
   The constant term is 4. Since $$ b^2 = 4 $$, we have $$ b = 2 $$.

3. **Check the middle term:**  
   The middle term should be $$ 2abx = 2 \times 1 \times 2x = 4x $$.  
   However, in the trinomial, the middle term is $$-4x$$, which suggests that $$ b = -2 $$.

Putting it all together:
$$
x^2 - 4x + 4 = (x - 2)^2
$$

**Factored Form:**
$$
(x - 2)^2
$$
The trinomial $$ x^{2} - 4x + 4 $$ factors to $$ (x - 2)^2 $$.

Factor the following trinomial.

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