Simplify ((x^2-12x+32)/(2x^3-18x^2+40x)*(x^2-1)/(x^2-7x-8))÷((x^2+4x-5)/(25-x^2))

$\frac{x^{2} - 12 x + 32}{2 x^{3} - 18 x^{2} + 40 x} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \div \frac{x^{2} + 4 x - 5}{25 - x^{2}}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 12 x + 32}{2 x^{3} - 18 x^{2} + 40 x} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $x^{2} - 12 x + 32$ 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 $32$ and whose sum is $- 12$ .

$- 8, - 4$

Write the factored form using these integers.

$\frac{(x - 8) (x - 4)}{2 x^{3} - 18 x^{2} + 40 x} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Simplify the denominator.

Factor $2 x$ out of $2 x^{3} - 18 x^{2} + 40 x$ .

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

$\frac{(x - 8) (x - 4)}{2 x (x^{2}) - 18 x^{2} + 40 x} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $2 x$ out of $- 18 x^{2}$ .

$\frac{(x - 8) (x - 4)}{2 x (x^{2}) + 2 x (- 9 x) + 40 x} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $2 x$ out of $40 x$ .

$\frac{(x - 8) (x - 4)}{2 x (x^{2}) + 2 x (- 9 x) + 2 x (20)} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $2 x$ out of $2 x (x^{2}) + 2 x (- 9 x)$ .

$\frac{(x - 8) (x - 4)}{2 x (x^{2} - 9 x) + 2 x (20)} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $2 x$ out of $2 x (x^{2} - 9 x) + 2 x (20)$ .

$\frac{(x - 8) (x - 4)}{2 x (x^{2} - 9 x + 20)} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $x^{2} - 9 x + 20$ 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 $20$ and whose sum is $- 9$ .

$- 5, - 4$

Write the factored form using these integers.

$\frac{(x - 8) (x - 4)}{2 x ((x - 5) (x - 4))} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

$\frac{(x - 8) (x - 4)}{2 x (x - 5) (x - 4)} \cdot \frac{x^{2} - 1}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Simplify the numerator.

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

$\frac{(x - 8) (x - 4)}{2 x (x - 5) (x - 4)} \cdot \frac{x^{2} - 1^{2}}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

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 $b = 1$ .

$\frac{(x - 8) (x - 4)}{2 x (x - 5) (x - 4)} \cdot \frac{(x + 1) (x - 1)}{x^{2} - 7 x - 8} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Factor $x^{2} - 7 x - 8$ 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 $- 7$ .

$- 8, 1$

Write the factored form using these integers.

$\frac{(x - 8) (x - 4)}{2 x (x - 5) (x - 4)} \cdot \frac{(x + 1) (x - 1)}{(x - 8) (x + 1)} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Simplify terms.

Cancel the common factor of $x - 8$ .

Cancel the common factor.

$\frac{(x - 8) (x - 4)}{2 x (x - 5) (x - 4)} \cdot \frac{(x + 1) (x - 1)}{(x - 8) (x + 1)} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Rewrite the expression.

$\frac{x - 4}{2 x (x - 5) (x - 4)} \cdot \frac{(x + 1) (x - 1)}{x + 1} \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Multiply $\frac{x - 4}{2 x (x - 5) (x - 4)}$ by $\frac{(x + 1) (x - 1)}{x + 1}$ .

$\frac{(x - 4) ((x + 1) (x - 1))}{2 x (x - 5) (x - 4) (x + 1)} \cdot \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{(x - 4) ((x + 1) (x - 1))}{2 x (x - 5) (x - 4) (x + 1)} \cdot \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Rewrite the expression.

$\frac{(x + 1) (x - 1)}{2 x (x - 5) (x + 1)} \cdot \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{(x + 1) (x - 1)}{2 x (x - 5) (x + 1)} \cdot \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Rewrite the expression.

$\frac{x - 1}{2 x (x - 5)} \cdot \frac{25 - x^{2}}{x^{2} + 4 x - 5}$

Simplify the numerator.

Rewrite $25$ as $5^{2}$ .

$\frac{x - 1}{2 x (x - 5)} \cdot \frac{5^{2} - x^{2}}{x^{2} + 4 x - 5}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$\frac{x - 1}{2 x (x - 5)} \cdot \frac{(5 + x) (5 - x)}{x^{2} + 4 x - 5}$

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$\frac{x - 1}{2 x (x - 5)} \cdot \frac{(5 + x) (5 - x)}{(x - 1) (x + 5)}$

Simplify terms.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{x - 1}{2 x (x - 5)} \cdot \frac{(5 + x) (5 - x)}{(x - 1) (x + 5)}$

Rewrite the expression.

$\frac{1}{2 x (x - 5)} \cdot \frac{(5 + x) (5 - x)}{x + 5}$

Multiply $\frac{1}{2 x (x - 5)}$ by $\frac{(5 + x) (5 - x)}{x + 5}$ .

$\frac{(5 + x) (5 - x)}{2 x (x - 5) (x + 5)}$

Cancel the common factor of $5 + x$ and $x + 5$ .

Reorder terms.

$\frac{(x + 5) (5 - x)}{2 x (x - 5) (x + 5)}$

Cancel the common factor.

$\frac{(x + 5) (5 - x)}{2 x (x - 5) (x + 5)}$

Rewrite the expression.

$\frac{5 - x}{2 x (x - 5)}$

Cancel the common factor of $5 - x$ and $x - 5$ .

Rewrite $5$ as $- 1 (- 5)$ .

$\frac{- 1 (- 5) - x}{2 x (x - 5)}$

Factor $- 1$ out of $- x$ .

$\frac{- 1 (- 5) - (x)}{2 x (x - 5)}$

Factor $- 1$ out of $- 1 (- 5) - (x)$ .

$\frac{- 1 (- 5 + x)}{2 x (x - 5)}$

Reorder terms.

$\frac{- 1 (x - 5)}{2 x (x - 5)}$

Cancel the common factor.

$\frac{- 1 (x - 5)}{2 x (x - 5)}$

Rewrite the expression.

$\frac{- 1}{2 x}$

Move the negative in front of the fraction.

$- \frac{1}{2 x}$

Simplify ((x^2-12x+32)/(2x^3-18x^2+40x)*(x^2-1)/(x^2-7x-8))÷((x^2+4x-5)/(25-x^2))

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