Q. \(x^2-2x+1\)

Q: Find x values where x^2-2x+1 \ge 0.

Since (x-1)^2 \ge 0 for all real x, the inequality holds for all real x.

Answer

\(x^2-2x+1\) is a perfect square.

\[
x^2-2x+1=(x-1)^2
\]

\(\boxed{(x-1)^2}\)

Detailed Explanation

We want to simplify the expression \(x^2 – 2x + 1\). A common method for this type of quadratic is to factor it as a perfect square.

Step 1: Recognize the pattern

A perfect square trinomial has the form:

\[
a^2 – 2ab + b^2
\]

Notice that our expression matches this structure:

\(x^2\) corresponds to \(a^2\),

\(-2x\) corresponds to \(-2ab\),

\(1\) corresponds to \(b^2\).

Step 2: Identify \(a\) and \(b\)

Let \(a = x\) and \(b = 1\). Then:

\[
a^2 = x^2
\]
\[
-2ab = -2(x)(1) = -2x
\]
\[
b^2 = 1^2 = 1
\]

These match the terms in the given expression exactly.

Step 3: Rewrite as a perfect square

So \(x^2 – 2x + 1\) factors as:

\[
x^2 – 2x + 1 = (x – 1)^2
\]

Final Answer

\[
x^2 – 2x + 1 = (x – 1)^2
\]

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Algebra FAQ

What is the factorization of \(x^2-2x+1\)?

\[x^2-2x+1=(x-1)^2\]

What is the vertex (and minimum value) of \(x^2-2x+1\)?

\[x^2-2x+1=(x-1)^2 \Rightarrow \text{vertex }(1,0),\ \text{minimum }0\]

Solve \(x^2-2x+1=0\).

\[(x-1)^2=0 \Rightarrow x=1\]

Expand \((x-1)^2\) to verify it matches \(x^2-2x+1\).

\[(x-1)^2=x^2-2x+1\]

Find \(x\) values where \(x^2-2x+1 \ge 0\).

Since \((x-1)^2 \ge 0\) for all real \(x\), the inequality holds for all real \(x\).

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