Simplify ((7x-14)/(x^2-3x-4)*(x^2-16)/(2x^2-3x-2))÷((x+4)/(4x^2-1))

$\frac{7 x - 14}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 16}{2 x^{2} - 3 x - 2} \div \frac{x + 4}{4 x^{2} - 1}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{7 x - 14}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 16}{2 x^{2} - 3 x - 2} \frac{4 x^{2} - 1}{x + 4}$

Step 2

Factor $7$ out of $7 x - 14$ .

Factor $7$ out of $7 x$ .

$\frac{7 (x) - 14}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 16}{2 x^{2} - 3 x - 2} \frac{4 x^{2} - 1}{x + 4}$

Factor $7$ out of $- 14$ .

$\frac{7 x + 7 \cdot - 2}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 16}{2 x^{2} - 3 x - 2} \frac{4 x^{2} - 1}{x + 4}$

Factor $7$ out of $7 x + 7 \cdot - 2$ .

$\frac{7 (x - 2)}{x^{2} - 3 x - 4} \cdot \frac{x^{2} - 16}{2 x^{2} - 3 x - 2} \frac{4 x^{2} - 1}{x + 4}$

Step 3

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$\frac{7 (x - 2)}{(x - 4) (x + 1)} \cdot \frac{x^{2} - 16}{2 x^{2} - 3 x - 2} \frac{4 x^{2} - 1}{x + 4}$

Step 4

Simplify the numerator.

Rewrite $16$ as $4^{2}$ .

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

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 $b = 4$ .

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

Step 5

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 $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = - 3$ .

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

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

Rewrite $- 3$ as $1$ plus $- 4$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

Step 6

Simplify terms.

Cancel the common factor of $x - 2$ .

Factor $x - 2$ out of $7 (x - 2)$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 4$ .

Factor $x - 4$ out of $(x + 4) (x - 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{7}{x + 1} \cdot \frac{x + 4}{2 x + 1} \frac{4 x^{2} - 1}{x + 4}$

Multiply $\frac{7}{x + 1}$ by $\frac{x + 4}{2 x + 1}$ .

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

Cancel the common factor of $x + 4$ .

Factor $x + 4$ out of $7 (x + 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{7}{(x + 1) (2 x + 1)} (4 x^{2} - 1)$

Multiply $\frac{7}{(x + 1) (2 x + 1)}$ by $4 x^{2} - 1$ .

$\frac{7 (4 x^{2} - 1)}{(x + 1) (2 x + 1)}$

Step 7

Simplify the numerator.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

Step 8

Cancel the common factor of $2 x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify ((7x-14)/(x^2-3x-4)*(x^2-16)/(2x^2-3x-2))÷((x+4)/(4x^2-1))

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