Divide ((x^2-9)/(x+1))÷((x^2-6x+9)/(7x+7))

$\frac{x^{2} - 9}{x + 1} \div \frac{x^{2} - 6 x + 9}{7 x + 7}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 9}{x + 1} \cdot \frac{7 x + 7}{x^{2} - 6 x + 9}$

Step 2

Simplify the numerator.

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

$\frac{x^{2} - 3^{2}}{x + 1} \cdot \frac{7 x + 7}{x^{2} - 6 x + 9}$

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 $b = 3$ .

$\frac{(x + 3) (x - 3)}{x + 1} \cdot \frac{7 x + 7}{x^{2} - 6 x + 9}$

Step 3

Factor $7$ out of $7 x + 7$ .

Factor $7$ out of $7 x$ .

$\frac{(x + 3) (x - 3)}{x + 1} \cdot \frac{7 (x) + 7}{x^{2} - 6 x + 9}$

Factor $7$ out of $7$ .

$\frac{(x + 3) (x - 3)}{x + 1} \cdot \frac{7 (x) + 7 (1)}{x^{2} - 6 x + 9}$

Factor $7$ out of $7 (x) + 7 (1)$ .

$\frac{(x + 3) (x - 3)}{x + 1} \cdot \frac{7 (x + 1)}{x^{2} - 6 x + 9}$

Step 4

Factor using the perfect square rule.

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

$\frac{(x + 3) (x - 3)}{x + 1} \cdot \frac{7 (x + 1)}{x^{2} - 6 x + 3^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Step 5

Cancel the common factor of $x - 3$ .

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

Cancel the common factor of $x + 1$ .

Factor $x + 1$ out of $7 (x + 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 7

Multiply $x + 3$ by $\frac{7}{x - 3}$ .

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

Step 8

Move $7$ to the left of $x + 3$ .

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

Divide ((x^2-9)/(x+1))÷((x^2-6x+9)/(7x+7))

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