Simplify ((6x+6)/(x-1))÷((x^2+3x+2)/(2x-2))

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

Step 1

To divide by a fraction, multiply by its reciprocal.

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

Step 2

Simplify with factoring out.

Factor $6$ out of $6 x + 6$ .

Factor $6$ out of $6 x$ .

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

Factor $6$ out of $6$ .

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

Factor $6$ out of $6 (x) + 6 (1)$ .

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

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

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (x) + 2 (- 1)$ .

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

Step 3

Factor $x^{2} + 3 x + 2$ 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 $2$ and whose sum is $3$ .

$1, 2$

Write the factored form using these integers.

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

Step 4

Simplify terms.

Cancel the common factor of $x + 1$ .

Factor $x + 1$ out of $6 (x + 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

Factor $x - 1$ out of $2 (x - 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$6 \frac{2}{x + 2}$

Combine $6$ and $\frac{2}{x + 2}$ .

$\frac{6 \cdot 2}{x + 2}$

Multiply $6$ by $2$ .

$\frac{12}{x + 2}$

Simplify ((6x+6)/(x-1))÷((x^2+3x+2)/(2x-2))

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