Simplify ((6x^2-24)/(x^2+9x+14))÷((3x^2+6x-24)/(x^2+2x-35))

$\frac{6 x^{2} - 24}{x^{2} + 9 x + 14} \div \frac{3 x^{2} + 6 x - 24}{x^{2} + 2 x - 35}$

To divide by a fraction, multiply by its reciprocal.

$\frac{6 x^{2} - 24}{x^{2} + 9 x + 14} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Simplify the numerator.

Factor $6$ out of $6 x^{2} - 24$ .

Factor $6$ out of $6 x^{2}$ .

$\frac{6 (x^{2}) - 24}{x^{2} + 9 x + 14} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Factor $6$ out of $- 24$ .

$\frac{6 x^{2} + 6 \cdot - 4}{x^{2} + 9 x + 14} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Factor $6$ out of $6 x^{2} + 6 \cdot - 4$ .

$\frac{6 (x^{2} - 4)}{x^{2} + 9 x + 14} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Rewrite $4$ as $2^{2}$ .

$\frac{6 (x^{2} - 2^{2})}{x^{2} + 9 x + 14} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

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 $b = 2$ .

$\frac{6 (x + 2) (x - 2)}{x^{2} + 9 x + 14} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Factor $x^{2} + 9 x + 14$ 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 $14$ and whose sum is $9$ .

$2, 7$

Write the factored form using these integers.

$\frac{6 (x + 2) (x - 2)}{(x + 2) (x + 7)} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Cancel the common factor of $x + 2$ .

Cancel the common factor.

$\frac{6 (x + 2) (x - 2)}{(x + 2) (x + 7)} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Rewrite the expression.

$\frac{6 (x - 2)}{x + 7} \cdot \frac{x^{2} + 2 x - 35}{3 x^{2} + 6 x - 24}$

Factor $x^{2} + 2 x - 35$ 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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 x^{2} + 6 x - 24}$

Simplify the denominator.

Factor $3$ out of $3 x^{2} + 6 x - 24$ .

Factor $3$ out of $3 x^{2}$ .

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x^{2}) + 6 x - 24}$

Factor $3$ out of $6 x$ .

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x^{2}) + 3 (2 x) - 24}$

Factor $3$ out of $- 24$ .

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 x^{2} + 3 (2 x) + 3 \cdot - 8}$

Factor $3$ out of $3 x^{2} + 3 (2 x)$ .

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x^{2} + 2 x) + 3 \cdot - 8}$

Factor $3$ out of $3 (x^{2} + 2 x) + 3 \cdot - 8$ .

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x^{2} + 2 x - 8)}$

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

$- 2, 4$

Write the factored form using these integers.

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 ((x - 2) (x + 4))}$

$\frac{6 (x - 2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x - 2) (x + 4)}$

Simplify terms.

Cancel the common factor of $3 (x - 2)$ .

Factor $3 (x - 2)$ out of $6 (x - 2)$ .

$\frac{3 (x - 2) (2)}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x - 2) (x + 4)}$

Cancel the common factor.

$\frac{3 (x - 2) \cdot 2}{x + 7} \cdot \frac{(x - 5) (x + 7)}{3 (x - 2) (x + 4)}$

Rewrite the expression.

$\frac{2}{x + 7} \cdot \frac{(x - 5) (x + 7)}{x + 4}$

Cancel the common factor of $x + 7$ .

Factor $x + 7$ out of $(x - 5) (x + 7)$ .

$\frac{2}{x + 7} \cdot \frac{(x + 7) (x - 5)}{x + 4}$

Cancel the common factor.

$\frac{2}{x + 7} \cdot \frac{(x + 7) (x - 5)}{x + 4}$

Rewrite the expression.

$2 \frac{x - 5}{x + 4}$

Combine $2$ and $\frac{x - 5}{x + 4}$ .

$\frac{2 (x - 5)}{x + 4}$

Simplify ((6x^2-24)/(x^2+9x+14))÷((3x^2+6x-24)/(x^2+2x-35))

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