Divide ((9x^2-18x-7)/(2x-4))÷((9x^2-49)/(x-2))

$\frac{9 x^{2} - 18 x - 7}{2 x - 4} \div \frac{9 x^{2} - 49}{x - 2}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{9 x^{2} - 18 x - 7}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

Step 2

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 $a \cdot c = 9 \cdot - 7 = - 63$ and whose sum is $b = - 18$ .

Factor $- 18$ out of $- 18 x$ .

$\frac{9 x^{2} - 18 (x) - 7}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

Rewrite $- 18$ as $3$ plus $- 21$

$\frac{9 x^{2} + (3 - 21) x - 7}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

Apply the distributive property.

$\frac{9 x^{2} + 3 x - 21 x - 7}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(9 x^{2} + 3 x) - 21 x - 7}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

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

$\frac{3 x (3 x + 1) - 7 (3 x + 1)}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

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

$\frac{(3 x + 1) (3 x - 7)}{2 x - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

Step 3

Factor $2$ out of $2 x - 4$ .

Factor $2$ out of $2 x$ .

$\frac{(3 x + 1) (3 x - 7)}{2 (x) - 4} \cdot \frac{x - 2}{9 x^{2} - 49}$

Factor $2$ out of $- 4$ .

$\frac{(3 x + 1) (3 x - 7)}{2 x + 2 \cdot - 2} \cdot \frac{x - 2}{9 x^{2} - 49}$

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

$\frac{(3 x + 1) (3 x - 7)}{2 (x - 2)} \cdot \frac{x - 2}{9 x^{2} - 49}$

Step 4

Cancel the common factor of $x - 2$ .

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

$\frac{(3 x + 1) (3 x - 7)}{(x - 2) \cdot 2} \cdot \frac{x - 2}{9 x^{2} - 49}$

Cancel the common factor.

$\frac{(3 x + 1) (3 x - 7)}{(x - 2) \cdot 2} \cdot \frac{x - 2}{9 x^{2} - 49}$

Rewrite the expression.

$\frac{(3 x + 1) (3 x - 7)}{2} \cdot \frac{1}{9 x^{2} - 49}$

Step 5

Multiply $\frac{(3 x + 1) (3 x - 7)}{2}$ by $\frac{1}{9 x^{2} - 49}$ .

$\frac{(3 x + 1) (3 x - 7)}{2 (9 x^{2} - 49)}$

Step 6

Simplify the denominator.

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

$\frac{(3 x + 1) (3 x - 7)}{2 ({(3 x)}^{2} - 49)}$

Rewrite $49$ as $7^{2}$ .

$\frac{(3 x + 1) (3 x - 7)}{2 ({(3 x)}^{2} - 7^{2})}$

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 $b = 7$ .

$\frac{(3 x + 1) (3 x - 7)}{2 (3 x + 7) (3 x - 7)}$

Step 7

Cancel the common factor of $3 x - 7$ .

Cancel the common factor.

$\frac{(3 x + 1) (3 x - 7)}{2 (3 x + 7) (3 x - 7)}$

Rewrite the expression.

$\frac{3 x + 1}{2 (3 x + 7)}$

Divide ((9x^2-18x-7)/(2x-4))÷((9x^2-49)/(x-2))

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