Pre-Algebra Examples

$\frac{x^{2} - 6 x - 27}{3 x^{2} - 243} \div \frac{x^{2} + 12 x + 27}{x^{2} + 18 x + 81}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 6 x - 27}{3 x^{2} - 243} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 2

Factor $x^{2} - 6 x - 27$ using the AC method.

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Step 2.1

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 $- 27$ and whose sum is $- 6$ .

$- 9, 3$

Step 2.2

Write the factored form using these integers.

$\frac{(x - 9) (x + 3)}{3 x^{2} - 243} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 3

Simplify the denominator.

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Step 3.1

Factor $3$ out of $3 x^{2} - 243$ .

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Step 3.1.1

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

$\frac{(x - 9) (x + 3)}{3 (x^{2}) - 243} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 3.1.2

Factor $3$ out of $- 243$ .

$\frac{(x - 9) (x + 3)}{3 x^{2} + 3 \cdot - 81} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 3.1.3

Factor $3$ out of $3 x^{2} + 3 \cdot - 81$ .

$\frac{(x - 9) (x + 3)}{3 (x^{2} - 81)} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 3.2

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

$\frac{(x - 9) (x + 3)}{3 (x^{2} - 9^{2})} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 3.3

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 $b = 9$ .

$\frac{(x - 9) (x + 3)}{3 (x + 9) (x - 9)} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 4

Cancel the common factor of $x - 9$ .

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Step 4.1

Cancel the common factor.

$\frac{(x - 9) (x + 3)}{3 (x + 9) (x - 9)} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 4.2

Rewrite the expression.

$\frac{x + 3}{3 (x + 9)} \cdot \frac{x^{2} + 18 x + 81}{x^{2} + 12 x + 27}$

Step 5

Factor using the perfect square rule.

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Step 5.1

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

$\frac{x + 3}{3 (x + 9)} \cdot \frac{x^{2} + 18 x + 9^{2}}{x^{2} + 12 x + 27}$

Step 5.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 x = 2 \cdot x \cdot 9$

Step 5.3

Rewrite the polynomial.

$\frac{x + 3}{3 (x + 9)} \cdot \frac{x^{2} + 2 \cdot x \cdot 9 + 9^{2}}{x^{2} + 12 x + 27}$

Step 5.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 9$ .

$\frac{x + 3}{3 (x + 9)} \cdot \frac{{(x + 9)}^{2}}{x^{2} + 12 x + 27}$

Step 6

Factor $x^{2} + 12 x + 27$ using the AC method.

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Step 6.1

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 $27$ and whose sum is $12$ .

$3, 9$

Step 6.2

Write the factored form using these integers.

$\frac{x + 3}{3 (x + 9)} \cdot \frac{{(x + 9)}^{2}}{(x + 3) (x + 9)}$

Step 7

Cancel the common factor of $x + 3$ .

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Step 7.1

Cancel the common factor.

$\frac{x + 3}{3 (x + 9)} \cdot \frac{{(x + 9)}^{2}}{(x + 3) (x + 9)}$

Step 7.2

Rewrite the expression.

$\frac{1}{3 (x + 9)} \cdot \frac{{(x + 9)}^{2}}{x + 9}$

Step 8

Cancel the common factor of $x + 9$ .

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Step 8.1

Factor $x + 9$ out of $3 (x + 9)$ .

$\frac{1}{(x + 9) \cdot 3} \cdot \frac{{(x + 9)}^{2}}{x + 9}$

Step 8.2

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

$\frac{1}{(x + 9) \cdot 3} \cdot \frac{(x + 9) (x + 9)}{x + 9}$

Step 8.3

Cancel the common factor.

$\frac{1}{(x + 9) \cdot 3} \cdot \frac{(x + 9) (x + 9)}{x + 9}$

Step 8.4

Rewrite the expression.

$\frac{1}{3} \cdot \frac{x + 9}{x + 9}$

Step 9

Multiply $\frac{1}{3}$ by $\frac{x + 9}{x + 9}$ .

$\frac{x + 9}{3 (x + 9)}$

Step 10

Cancel the common factor of $x + 9$ .

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Step 10.1

Cancel the common factor.

$\frac{x + 9}{3 (x + 9)}$

Step 10.2

Rewrite the expression.

$\frac{1}{3}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{3}$

Decimal Form:

$0.\overline{3}$