Factor. w^2 - 2w + 1 ______

Check for Perfect Square Trinomial: Determine if the quadratic can be factored as a perfect square trinomial. A perfect square trinomial is in the form (a - b)^2 = a^2 - 2ab + b^2. We can compare w^2 - 2w + 1 to the form a^2 - 2ab + b^2 to see if it matches.Identify a and b: Identify a and b in the expression w^2 - 2w + 1. Here, a = w and b = 1 because (w)^2 = w^2 and (1)^2 = 1. The middle term -2w should be equal to -2ab, which is -2(w)(1) = -2w.Confirm Trinomial Type: Confirm that the expression is a perfect square trinomial. Since w^2 is the square of w, 1 is the square of 1, and -2w is twice the product of w and 1, the expression w^2 - 2w + 1 is indeed a perfect square trinomial.Factor Using Formula: Factor the expression using the perfect square trinomial formula. The factored form of w^2 - 2w + 1 is (w - 1)^2 because (w - 1)(w - 1) gives w^2 - w - w + 1, which simplifies to w^2 - 2w + 1.

Home

Math Problems

Algebra 1

Factor quadratics: special cases

Full solution

Q. Factor.

\newline

w^2 - 2w + 1

Check for Perfect Square Trinomial: Determine if the quadratic can be factored as a perfect square trinomial. A perfect square trinomial is in the form $(a - b)^2 = a^2 - 2ab + b^2$ . We can compare $w^2 - 2w + 1$ to the form $a^2 - 2ab + b^2$ to see if it matches.
Identify $a$ and $b$ : Identify $a$ and $b$ in the expression $w^2 - 2w + 1$ . Here, $a = w$ and $b = 1$ because $(w)^2 = w^2$ and $(1)^2 = 1$ . The middle term $-2w$ should be equal to $b$ $0$ , which is $b$ $1$ .
Confirm Trinomial Type: Confirm that the expression is a perfect square trinomial. $\newline$ Since $w^2$ is the square of $w$ , $1$ is the square of $1$ , and $-2w$ is twice the product of $w$ and $1$ , the expression $w^2 - 2w + 1$ is indeed a perfect square trinomial.
Factor Using Formula: Factor the expression using the perfect square trinomial formula. $\newline$ The factored form of $w^2 - 2w + 1$ is $(w - 1)^2$ because $(w - 1)(w - 1)$ gives $w^2 - w - w + 1$ , which simplifies to $w^2 - 2w + 1$ .