Divide ((x^2-5x-36)/(x+2))÷(x^2-18x+81)

$\frac{x^{2} - 5 x - 36}{x + 2} \div (x^{2} - 18 x + 81)$

Step 1

Rewrite the division as a fraction.

$\frac{\frac{x^{2} - 5 x - 36}{x + 2}}{x^{2} - 18 x + 81}$

Step 2

Multiply the numerator by the reciprocal of the denominator.

$\frac{x^{2} - 5 x - 36}{x + 2} \cdot \frac{1}{x^{2} - 18 x + 81}$

Step 3

Factor $x^{2} - 5 x - 36$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 36$ and whose sum is $- 5$ .

$- 9, 4$

Write the factored form using these integers.

$\frac{(x - 9) (x + 4)}{x + 2} \cdot \frac{1}{x^{2} - 18 x + 81}$

Step 4

Factor using the perfect square rule.

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

$\frac{(x - 9) (x + 4)}{x + 2} \cdot \frac{1}{x^{2} - 18 x + 9^{2}}$

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.

$\frac{(x - 9) (x + 4)}{x + 2} \cdot \frac{1}{x^{2} - 2 \cdot x \cdot 9 + 9^{2}}$

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

$\frac{(x - 9) (x + 4)}{x + 2} \cdot \frac{1}{{(x - 9)}^{2}}$

Step 5

Cancel the common factor of $x - 9$ .

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

$\frac{(x - 9) (x + 4)}{x + 2} \cdot \frac{1}{(x - 9) (x - 9)}$

Cancel the common factor.

$\frac{(x - 9) (x + 4)}{x + 2} \cdot \frac{1}{(x - 9) (x - 9)}$

Rewrite the expression.

$\frac{x + 4}{x + 2} \cdot \frac{1}{x - 9}$

Step 6

Multiply $\frac{x + 4}{x + 2}$ by $\frac{1}{x - 9}$ .

$\frac{x + 4}{(x + 2) (x - 9)}$

Step 7

Split the fraction $\frac{x + 4}{(x + 2) (x - 9)}$ into two fractions.

$\frac{x}{(x + 2) (x - 9)} + \frac{4}{(x + 2) (x - 9)}$

Divide ((x^2-5x-36)/(x+2))÷(x^2-18x+81)

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