Pre-Algebra Examples

$\frac{x^{2} - 81}{x^{2}} \cdot \frac{x^{2} - 9 x}{x^{2} + x - 90}$

Step 1

Rewrite $81$ as $9^{2}$ .

$\frac{x^{2} - 9^{2}}{x^{2}} \cdot \frac{x^{2} - 9 x}{x^{2} + x - 90}$

Step 2

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 $b = 9$ .

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x^{2} - 9 x}{x^{2} + x - 90}$

Step 3

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

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Step 3.1

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

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x \cdot x - 9 x}{x^{2} + x - 90}$

Step 3.2

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

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x \cdot x + x \cdot - 9}{x^{2} + x - 90}$

Step 3.3

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

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x (x - 9)}{x^{2} + x - 90}$

Step 4

Factor $x^{2} + x - 90$ using the AC method.

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Step 4.1

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 $- 90$ and whose sum is $1$ .

$- 9, 10$

Step 4.2

Write the factored form using these integers.

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x (x - 9)}{(x - 9) (x + 10)}$

Step 5

Reduce the expression $\frac{x (x - 9)}{(x - 9) (x + 10)}$ by cancelling the common factors.

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Step 5.1

Cancel the common factor.

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x (x - 9)}{(x - 9) (x + 10)}$

Step 5.2

Rewrite the expression.

$\frac{(x + 9) (x - 9)}{x^{2}} \cdot \frac{x}{x + 10}$