Simplify (2x-10)/(x+8)*(x^2-64)/(x^2+3x-40)

$\frac{2 x - 10}{x + 8} \cdot \frac{x^{2} - 64}{x^{2} + 3 x - 40}$

Step 1

Factor $2$ out of $2 x - 10$ .

Factor $2$ out of $2 x$ .

$\frac{2 (x) - 10}{x + 8} \cdot \frac{x^{2} - 64}{x^{2} + 3 x - 40}$

Factor $2$ out of $- 10$ .

$\frac{2 x + 2 \cdot - 5}{x + 8} \cdot \frac{x^{2} - 64}{x^{2} + 3 x - 40}$

Factor $2$ out of $2 x + 2 \cdot - 5$ .

$\frac{2 (x - 5)}{x + 8} \cdot \frac{x^{2} - 64}{x^{2} + 3 x - 40}$

Step 2

Simplify the numerator.

Rewrite $64$ as $8^{2}$ .

$\frac{2 (x - 5)}{x + 8} \cdot \frac{x^{2} - 8^{2}}{x^{2} + 3 x - 40}$

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 $b = 8$ .

$\frac{2 (x - 5)}{x + 8} \cdot \frac{(x + 8) (x - 8)}{x^{2} + 3 x - 40}$

Step 3

Factor $x^{2} + 3 x - 40$ 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 $- 40$ and whose sum is $3$ .

$- 5, 8$

Write the factored form using these integers.

$\frac{2 (x - 5)}{x + 8} \cdot \frac{(x + 8) (x - 8)}{(x - 5) (x + 8)}$

Step 4

Simplify terms.

Cancel the common factor of $x - 5$ .

Factor $x - 5$ out of $2 (x - 5)$ .

$\frac{(x - 5) \cdot 2}{x + 8} \cdot \frac{(x + 8) (x - 8)}{(x - 5) (x + 8)}$

Cancel the common factor.

$\frac{(x - 5) \cdot 2}{x + 8} \cdot \frac{(x + 8) (x - 8)}{(x - 5) (x + 8)}$

Rewrite the expression.

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

Cancel the common factor of $x + 8$ .

Cancel the common factor.

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

Rewrite the expression.

$2 \cdot \frac{x - 8}{x + 8}$

Combine $2$ and $\frac{x - 8}{x + 8}$ .

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

Simplify (2x-10)/(x+8)*(x^2-64)/(x^2+3x-40)

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