Simplify (x^2+6x-7)/(2x^3+2x^2-4x)4/(x+7)(x^3-4x)/(6x^2-12x)

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x^{3} - 4 x}{6 x^{2} - 12 x}$

Step 1

Simplify the numerator.

Factor $x$ out of $x^{3} - 4 x$ .

Factor $x$ out of $x^{3}$ .

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x \cdot x^{2} - 4 x}{6 x^{2} - 12 x}$

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

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x \cdot x^{2} + x \cdot - 4}{6 x^{2} - 12 x}$

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

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x^{2} - 4)}{6 x^{2} - 12 x}$

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

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x^{2} - 2^{2})}{6 x^{2} - 12 x}$

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 $b = 2$ .

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x^{2} - 12 x}$

Step 2

Factor $6 x$ out of $6 x^{2} - 12 x$ .

Factor $6 x$ out of $6 x^{2}$ .

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x) - 12 x}$

Factor $6 x$ out of $- 12 x$ .

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x) + 6 x (- 2)}$

Factor $6 x$ out of $6 x (x) + 6 x (- 2)$ .

$\frac{x^{2} + 6 x - 7}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Step 3

Factor $x^{2} + 6 x - 7$ 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 $- 7$ and whose sum is $6$ .

$- 1, 7$

Write the factored form using these integers.

$\frac{(x - 1) (x + 7)}{2 x^{3} + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Step 4

Simplify the denominator.

Factor $2 x$ out of $2 x^{3} + 2 x^{2} - 4 x$ .

Factor $2 x$ out of $2 x^{3}$ .

$\frac{(x - 1) (x + 7)}{2 x (x^{2}) + 2 x^{2} - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Factor $2 x$ out of $2 x^{2}$ .

$\frac{(x - 1) (x + 7)}{2 x (x^{2}) + 2 x (x) - 4 x} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

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

$\frac{(x - 1) (x + 7)}{2 x (x^{2}) + 2 x (x) + 2 x (- 2)} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

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

$\frac{(x - 1) (x + 7)}{2 x (x^{2} + x) + 2 x (- 2)} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

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

$\frac{(x - 1) (x + 7)}{2 x (x^{2} + x - 2)} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Factor $x^{2} + x - 2$ 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 $1$ .

$- 1, 2$

Write the factored form using these integers.

$\frac{(x - 1) (x + 7)}{2 x ((x - 1) (x + 2))} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

$\frac{(x - 1) (x + 7)}{2 x (x - 1) (x + 2)} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Step 5

Simplify terms.

Cancel the common factor of $x + 7$ .

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

$\frac{(x + 7) (x - 1)}{2 x (x - 1) (x + 2)} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Cancel the common factor.

$\frac{(x + 7) (x - 1)}{2 x (x - 1) (x + 2)} \cdot \frac{4}{x + 7} \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Rewrite the expression.

$\frac{x - 1}{2 x (x - 1) (x + 2)} \cdot 4 \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Cancel the common factor of $2$ .

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

$\frac{x - 1}{2 ((x (x - 1)) (x + 2))} \cdot 4 \cdot \frac{x (x + 2) (x - 2)}{6 x (x - 2)}$

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine $\frac{x - 1}{(x (x - 1)) (x + 2)}$ and $2$ .

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

Cancel the common factor of $2$ .

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

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

Factor $2$ out of $6 x (x - 2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x (x + 2)$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{x - 1}{x - 1}$ by $\frac{x - 2}{3 x (x - 2)}$ .

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 2}{3 x (x - 2)}$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{x - 2}{3 x (x - 2)}$

Rewrite the expression.

$\frac{1}{3 x}$

Simplify (x^2+6x-7)/(2x^3+2x^2-4x)*4/(x+7)*(x^3-4x)/(6x^2-12x)

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Simplify (x^2+6x-7)/(2x^3+2x^2-4x)4/(x+7)(x^3-4x)/(6x^2-12x)