Divide ((2x^2-9x-5)/(x^2-8x+15))÷((18x+9)/(x^2-9))

$\frac{2 x^{2} - 9 x - 5}{x^{2} - 8 x + 15} \div \frac{18 x + 9}{x^{2} - 9}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{2 x^{2} - 9 x - 5}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$\frac{2 x^{2} - 9 (x) - 5}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Rewrite $- 9$ as $1$ plus $- 10$

$\frac{2 x^{2} + (1 - 10) x - 5}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Apply the distributive property.

$\frac{2 x^{2} + 1 x - 10 x - 5}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Multiply $x$ by $1$ .

$\frac{2 x^{2} + x - 10 x - 5}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{2} + x) - 10 x - 5}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Factor out the greatest common factor (GCF) from each group.

$\frac{x (2 x + 1) - 5 (2 x + 1)}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$\frac{(2 x + 1) (x - 5)}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 9}{18 x + 9}$

Step 3

Factor $x^{2} - 8 x + 15$ 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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

Write the factored form using these integers.

$\frac{(2 x + 1) (x - 5)}{(x - 5) (x - 3)} \cdot \frac{x^{2} - 9}{18 x + 9}$

Step 4

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$\frac{(2 x + 1) (x - 5)}{(x - 5) (x - 3)} \cdot \frac{x^{2} - 9}{18 x + 9}$

Rewrite the expression.

$\frac{2 x + 1}{x - 3} \cdot \frac{x^{2} - 9}{18 x + 9}$

Step 5

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

$\frac{2 x + 1}{x - 3} \cdot \frac{x^{2} - 3^{2}}{18 x + 9}$

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 $b = 3$ .

$\frac{2 x + 1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{18 x + 9}$

Step 6

Factor $9$ out of $18 x + 9$ .

Factor $9$ out of $18 x$ .

$\frac{2 x + 1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{9 (2 x) + 9}$

Factor $9$ out of $9$ .

$\frac{2 x + 1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{9 (2 x) + 9 (1)}$

Factor $9$ out of $9 (2 x) + 9 (1)$ .

$\frac{2 x + 1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{9 (2 x + 1)}$

Step 7

Cancel the common factor of $2 x + 1$ .

Factor $2 x + 1$ out of $9 (2 x + 1)$ .

$\frac{2 x + 1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{(2 x + 1) \cdot 9}$

Cancel the common factor.

$\frac{2 x + 1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{(2 x + 1) \cdot 9}$

Rewrite the expression.

$\frac{1}{x - 3} \cdot \frac{(x + 3) (x - 3)}{9}$

Step 8

Cancel the common factor of $x - 3$ .

Factor $x - 3$ out of $(x + 3) (x - 3)$ .

$\frac{1}{x - 3} \cdot \frac{(x - 3) (x + 3)}{9}$

Cancel the common factor.

$\frac{1}{x - 3} \cdot \frac{(x - 3) (x + 3)}{9}$

Rewrite the expression.

$\frac{x + 3}{9}$

Step 9

Split the fraction $\frac{x + 3}{9}$ into two fractions.

$\frac{x}{9} + \frac{3}{9}$

Step 10

Cancel the common factor of $3$ and $9$ .

Factor $3$ out of $3$ .

$\frac{x}{9} + \frac{3 (1)}{9}$

Cancel the common factors.

Factor $3$ out of $9$ .

$\frac{x}{9} + \frac{3 \cdot 1}{3 \cdot 3}$

Cancel the common factor.

$\frac{x}{9} + \frac{3 \cdot 1}{3 \cdot 3}$

Rewrite the expression.

$\frac{x}{9} + \frac{1}{3}$

Divide ((2x^2-9x-5)/(x^2-8x+15))÷((18x+9)/(x^2-9))

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