Simplify (x^2-64)/(x^2-9x)*(x+9)/(x^2+x-72)

$\frac{x^{2} - 64}{x^{2} - 9 x} \cdot \frac{x + 9}{x^{2} + x - 72}$

Simplify the numerator.

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

$\frac{x^{2} - 8^{2}}{x^{2} - 9 x} \cdot \frac{x + 9}{x^{2} + x - 72}$

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 $b = 8$ .

$\frac{(x + 8) (x - 8)}{x^{2} - 9 x} \cdot \frac{x + 9}{x^{2} + x - 72}$

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

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

$\frac{(x + 8) (x - 8)}{x \cdot x - 9 x} \cdot \frac{x + 9}{x^{2} + x - 72}$

Factor $x$ out of $- 9 x$ .

$\frac{(x + 8) (x - 8)}{x \cdot x + x \cdot - 9} \cdot \frac{x + 9}{x^{2} + x - 72}$

Factor $x$ out of $x \cdot x + x \cdot - 9$ .

$\frac{(x + 8) (x - 8)}{x (x - 9)} \cdot \frac{x + 9}{x^{2} + x - 72}$

Factor $x^{2} + x - 72$ 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 $- 72$ and whose sum is $1$ .

$- 8, 9$

Write the factored form using these integers.

$\frac{(x + 8) (x - 8)}{x (x - 9)} \cdot \frac{x + 9}{(x - 8) (x + 9)}$

Simplify terms.

Cancel the common factor of $x - 8$ .

Factor $x - 8$ out of $(x + 8) (x - 8)$ .

$\frac{(x - 8) (x + 8)}{x (x - 9)} \cdot \frac{x + 9}{(x - 8) (x + 9)}$

Cancel the common factor.

$\frac{(x - 8) (x + 8)}{x (x - 9)} \cdot \frac{x + 9}{(x - 8) (x + 9)}$

Rewrite the expression.

$\frac{x + 8}{x (x - 9)} \cdot \frac{x + 9}{x + 9}$

Multiply $\frac{x + 8}{x (x - 9)}$ by $\frac{x + 9}{x + 9}$ .

$\frac{(x + 8) (x + 9)}{x (x - 9) (x + 9)}$

Cancel the common factor of $x + 9$ .

Cancel the common factor.

$\frac{(x + 8) (x + 9)}{x (x - 9) (x + 9)}$

Rewrite the expression.

$\frac{x + 8}{x (x - 9)}$

Simplify (x^2-64)/(x^2-9x)*(x+9)/(x^2+x-72)

Related Questions

Support LCMGCF.com by disabling your adblocker.