Simplify ((x^2-3x-10)/(x+3))/((2x^2-7x-15)/(x^2-9))

$\frac{\frac{x^{2} - 3 x - 10}{x + 3}}{\frac{2 x^{2} - 7 x - 15}{x^{2} - 9}}$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$\frac{x^{2} - 3 x - 10}{x + 3} \cdot \frac{x^{2} - 9}{2 x^{2} - 7 x - 15}$

Step 2

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{x^{2} - 9}{2 x^{2} - 7 x - 15}$

Step 3

Simplify the numerator.

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

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{x^{2} - 3^{2}}{2 x^{2} - 7 x - 15}$

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 $b = 3$ .

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{(x + 3) (x - 3)}{2 x^{2} - 7 x - 15}$

Step 4

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 $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{(x + 3) (x - 3)}{2 x^{2} - 7 (x) - 15}$

Rewrite $- 7$ as $3$ plus $- 10$

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{(x + 3) (x - 3)}{2 x^{2} + (3 - 10) x - 15}$

Apply the distributive property.

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{(x + 3) (x - 3)}{2 x^{2} + 3 x - 10 x - 15}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 5) (x + 2)}{x + 3} \cdot \frac{(x + 3) (x - 3)}{(2 x^{2} + 3 x) - 10 x - 15}$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

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

Step 5

Cancel the common factor of $x - 5$ .

Factor $x - 5$ out of $(2 x + 3) (x - 5)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 7

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

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

Simplify ((x^2-3x-10)/(x+3))/((2x^2-7x-15)/(x^2-9))

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