Divide ((9x^2-3x-2)/(2x-6))÷((9x^2-4)/(x-3))

$\frac{9 x^{2} - 3 x - 2}{2 x - 6} \div \frac{9 x^{2} - 4}{x - 3}$

To divide by a fraction, multiply by its reciprocal.

$\frac{9 x^{2} - 3 x - 2}{2 x - 6} \cdot \frac{x - 3}{9 x^{2} - 4}$

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 $a \cdot c = 9 \cdot - 2 = - 18$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

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

Rewrite $- 3$ as $3$ plus $- 6$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Factor $2$ out of $2 x - 6$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $2 x + 2 \cdot - 3$ .

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

Cancel the common factor of $x - 3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{(3 x + 1) (3 x - 2)}{2}$ by $\frac{1}{9 x^{2} - 4}$ .

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

Simplify the denominator.

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

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

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

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

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 $b = 2$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide ((9x^2-3x-2)/(2x-6))÷((9x^2-4)/(x-3))

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