Evaluate the Function f(x)=(x^2-18x+81)/(x-9) , a=9

$f (x) = \frac{x^{2} - 18 x + 81}{x - 9}$ , $a = 9$

Replace the variable $a$ with $9$ in the expression.

$f (x) = \frac{x^{2} - 18 x + 81}{x - 9}$

Simplify the result.

Factor using the perfect square rule.

Rewrite $81$ as $9^{2}$ .

$f (x) = \frac{x^{2} - 18 x + 9^{2}}{x - 9}$

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

$18 x = 2 \cdot x \cdot 9$

Rewrite the polynomial.

$f (x) = \frac{x^{2} - 2 \cdot x \cdot 9 + 9^{2}}{x - 9}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 9$ .

$f (x) = \frac{{(x - 9)}^{2}}{x - 9}$

Cancel the common factor of ${(x - 9)}^{2}$ and $x - 9$ .

Factor $x - 9$ out of ${(x - 9)}^{2}$ .

$f (x) = \frac{(x - 9) (x - 9)}{x - 9}$

Cancel the common factors.

Multiply by $1$ .

$f (x) = \frac{(x - 9) (x - 9)}{(x - 9) \cdot 1}$

Cancel the common factor.

$f (x) = \frac{(x - 9) (x - 9)}{(x - 9) \cdot 1}$

Rewrite the expression.

$f (x) = \frac{x - 9}{1}$

Divide $x - 9$ by $1$ .

$f (x) = x - 9$

The final answer is $x - 9$ .

$x - 9$