Divide ((x^2-2x-15)/(2x^2+3x+1))÷((x^2-9)/(x^2-8x-9))

$\frac{x^{2} - 2 x - 15}{2 x^{2} + 3 x + 1} \div \frac{x^{2} - 9}{x^{2} - 8 x - 9}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 2 x - 15}{2 x^{2} + 3 x + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

Factor $x^{2} - 2 x - 15$ 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 $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$\frac{(x - 5) (x + 3)}{2 x^{2} + 3 x + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$\frac{(x - 5) (x + 3)}{2 x^{2} + 3 (x) + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

Rewrite $3$ as $1$ plus $2$

$\frac{(x - 5) (x + 3)}{2 x^{2} + (1 + 2) x + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

Apply the distributive property.

$\frac{(x - 5) (x + 3)}{2 x^{2} + 1 x + 2 x + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

Multiply $x$ by $1$ .

$\frac{(x - 5) (x + 3)}{2 x^{2} + x + 2 x + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 5) (x + 3)}{(2 x^{2} + x) + 2 x + 1} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

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

$\frac{(x - 5) (x + 3)}{x (2 x + 1) + 1 (2 x + 1)} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

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

$\frac{(x - 5) (x + 3)}{(2 x + 1) (x + 1)} \cdot \frac{x^{2} - 8 x - 9}{x^{2} - 9}$

Factor $x^{2} - 8 x - 9$ 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 $- 9$ and whose sum is $- 8$ .

$- 9, 1$

Write the factored form using these integers.

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

Simplify the denominator.

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

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

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 $b = 3$ .

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

Cancel the common factor of $x + 3$ .

Factor $x + 3$ out of $(x - 5) (x + 3)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 5}{2 x + 1} \cdot \frac{x - 9}{x - 3}$

Multiply $\frac{x - 5}{2 x + 1}$ by $\frac{x - 9}{x - 3}$ .

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

Divide ((x^2-2x-15)/(2x^2+3x+1))÷((x^2-9)/(x^2-8x-9))

Related Questions

Support LCMGCF.com by disabling your adblocker.