Divide ((x^2+x-12)/(9x^2-1))÷((x^2-5x+6)/(3x^2-5x-2))

$\frac{x^{2} + x - 12}{9 x^{2} - 1} \div \frac{x^{2} - 5 x + 6}{3 x^{2} - 5 x - 2}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} + x - 12}{9 x^{2} - 1} \cdot \frac{3 x^{2} - 5 x - 2}{x^{2} - 5 x + 6}$

Step 2

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$\frac{(x - 3) (x + 4)}{9 x^{2} - 1} \cdot \frac{3 x^{2} - 5 x - 2}{x^{2} - 5 x + 6}$

Step 3

Simplify the denominator.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$\frac{(x - 3) (x + 4)}{{(3 x)}^{2} - 1} \cdot \frac{3 x^{2} - 5 x - 2}{x^{2} - 5 x + 6}$

Rewrite $1$ as $1^{2}$ .

$\frac{(x - 3) (x + 4)}{{(3 x)}^{2} - 1^{2}} \cdot \frac{3 x^{2} - 5 x - 2}{x^{2} - 5 x + 6}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 1$ .

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{3 x^{2} - 5 x - 2}{x^{2} - 5 x + 6}$

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 $a \cdot c = 3 \cdot - 2 = - 6$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{3 x^{2} - 5 (x) - 2}{x^{2} - 5 x + 6}$

Rewrite $- 5$ as $1$ plus $- 6$

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{3 x^{2} + (1 - 6) x - 2}{x^{2} - 5 x + 6}$

Apply the distributive property.

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{3 x^{2} + 1 x - 6 x - 2}{x^{2} - 5 x + 6}$

Multiply $x$ by $1$ .

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{3 x^{2} + x - 6 x - 2}{x^{2} - 5 x + 6}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{(3 x^{2} + x) - 6 x - 2}{x^{2} - 5 x + 6}$

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

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{x (3 x + 1) - 2 (3 x + 1)}{x^{2} - 5 x + 6}$

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{(3 x + 1) (x - 2)}{x^{2} - 5 x + 6}$

Step 5

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{(3 x + 1) (x - 2)}{(x - 3) (x - 2)}$

Step 6

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{(x - 3) (x + 4)}{(3 x + 1) (3 x - 1)} \cdot \frac{(3 x + 1) (x - 2)}{(x - 3) (x - 2)}$

Rewrite the expression.

$\frac{x + 4}{(3 x + 1) (3 x - 1)} \cdot \frac{(3 x + 1) (x - 2)}{x - 2}$

Step 7

Cancel the common factor of $3 x + 1$ .

Cancel the common factor.

$\frac{x + 4}{(3 x + 1) (3 x - 1)} \cdot \frac{(3 x + 1) (x - 2)}{x - 2}$

Rewrite the expression.

$\frac{x + 4}{3 x - 1} \cdot \frac{x - 2}{x - 2}$

Step 8

Multiply $\frac{x + 4}{3 x - 1}$ by $\frac{x - 2}{x - 2}$ .

$\frac{(x + 4) (x - 2)}{(3 x - 1) (x - 2)}$

Step 9

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{(x + 4) (x - 2)}{(3 x - 1) (x - 2)}$

Rewrite the expression.

$\frac{x + 4}{3 x - 1}$

Divide ((x^2+x-12)/(9x^2-1))÷((x^2-5x+6)/(3x^2-5x-2))

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