Divide (x^2-81)/(9-x)

$\frac{x^{2} - 81}{9 - x}$

Step 1

Simplify the numerator.

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

$\frac{x^{2} - 9^{2}}{9 - x}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 9$ .

$\frac{(x + 9) (x - 9)}{9 - x}$

Step 2

Cancel the common factor of $x - 9$ and $9 - x$ .

Factor $- 1$ out of $x$ .

$\frac{(x + 9) (- 1 (- x) - 9)}{9 - x}$

Rewrite $- 9$ as $- 1 (9)$ .

$\frac{(x + 9) (- 1 (- x) - 1 (9))}{9 - x}$

Factor $- 1$ out of $- 1 (- x) - 1 (9)$ .

$\frac{(x + 9) (- 1 (- x + 9))}{9 - x}$

Reorder terms.

$\frac{(x + 9) (- 1 (- x + 9))}{- x + 9}$

Cancel the common factor.

$\frac{(x + 9) (- 1 (- x + 9))}{- x + 9}$

Divide $(x + 9) \cdot (- 1)$ by $1$ .

$(x + 9) \cdot (- 1)$

Step 3

Apply the distributive property.

$x \cdot - 1 + 9 \cdot - 1$

Step 4

Move $- 1$ to the left of $x$ .

$- 1 \cdot x + 9 \cdot - 1$

Step 5

Multiply $9$ by $- 1$ .

$- 1 \cdot x - 9$

Step 6

Rewrite $- 1 x$ as $- x$ .

$- x - 9$