Simplify ((x^2+6x+8)/(x+4))÷((x^2-4)/(4x-8))

$\frac{x^{2} + 6 x + 8}{x + 4} \div \frac{x^{2} - 4}{4 x - 8}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} + 6 x + 8}{x + 4} \cdot \frac{4 x - 8}{x^{2} - 4}$

Step 2

Factor $x^{2} + 6 x + 8$ 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 $8$ and whose sum is $6$ .

$2, 4$

Write the factored form using these integers.

$\frac{(x + 2) (x + 4)}{x + 4} \cdot \frac{4 x - 8}{x^{2} - 4}$

Step 3

Simplify terms.

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$\frac{(x + 2) (x + 4)}{x + 4} \cdot \frac{4 x - 8}{x^{2} - 4}$

Divide $x + 2$ by $1$ .

$(x + 2) \frac{4 x - 8}{x^{2} - 4}$

Factor $4$ out of $4 x - 8$ .

Factor $4$ out of $4 x$ .

$(x + 2) \frac{4 (x) - 8}{x^{2} - 4}$

Factor $4$ out of $- 8$ .

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

Factor $4$ out of $4 x + 4 \cdot - 2$ .

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

Step 4

Simplify the denominator.

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

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

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

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

Step 5

Reduce the expression by cancelling the common factors.

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{4 (x - 2)}{x - 2}$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{4 (x - 2)}{x - 2}$

Divide $4$ by $1$ .

$4$

Simplify ((x^2+6x+8)/(x+4))÷((x^2-4)/(4x-8))

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