Question

factor $x^2 - 18x + 81$.
the factored form is \boxed{}.

Explanation:

Step1: Identify the form

The quadratic expression $x^2 - 18x + 81$ is a perfect square trinomial. The general form of a perfect square trinomial is $a^2 - 2ab + b^2=(a - b)^2$.

Step2: Match with the general form

For $x^2 - 18x + 81$, we have $a = x$ (since $x^2=a^2$), and we need to find $b$ such that $2ab = 18x$. Substituting $a = x$, we get $2\times x\times b=18x$. Solving for $b$, we divide both sides by $2x$ (assuming $x
eq0$), so $b = \frac{18x}{2x}=9$. Also, $b^2 = 9^2 = 81$, which matches the constant term.

Step3: Factor the expression

Using the perfect square trinomial formula $a^2 - 2ab + b^2=(a - b)^2$ with $a = x$ and $b = 9$, we get $x^2 - 18x + 81=(x - 9)^2$.

Answer:

$(x - 9)^2$