Solve by Factoring x^2-2x+1=0

$x^{2} - 2 x + 1 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x^{2} - 2 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x - 1)}^{2} = 0$

Step 2

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 3

Add $1$ to both sides of the equation.

$x = 1$