Pre-Algebra Examples

$\frac{\frac{b^{2} - 6 b + 9}{b^{2} - b - 6}}{b^{2} - \frac{9}{4}}$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$\frac{b^{2} - 6 b + 9}{b^{2} - b - 6} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $9$ as $3^{2}$ .

$\frac{b^{2} - 6 b + 3^{2}}{b^{2} - b - 6} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 2.2

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

$6 b = 2 \cdot b \cdot 3$

Step 2.3

Rewrite the polynomial.

$\frac{b^{2} - 2 \cdot b \cdot 3 + 3^{2}}{b^{2} - b - 6} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = b$ and $b = 3$ .

$\frac{{(b - 3)}^{2}}{b^{2} - b - 6} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 3

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

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Step 3.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 3.2

Write the factored form using these integers.

$\frac{{(b - 3)}^{2}}{(b - 3) (b + 2)} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 4

Cancel the common factor of ${(b - 3)}^{2}$ and $b - 3$ .

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Step 4.1

Factor $b - 3$ out of ${(b - 3)}^{2}$ .

$\frac{(b - 3) (b - 3)}{(b - 3) (b + 2)} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Cancel the common factor.

$\frac{(b - 3) (b - 3)}{(b - 3) (b + 2)} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 4.2.2

Rewrite the expression.

$\frac{b - 3}{b + 2} \cdot \frac{1}{b^{2} - \frac{9}{4}}$

Step 5

Simplify the denominator.

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Step 5.1

To write $b^{2}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{b^{2} \cdot \frac{4}{4} - \frac{9}{4}}$

Step 5.2

Combine $b^{2}$ and $\frac{4}{4}$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{b^{2} \cdot 4}{4} - \frac{9}{4}}$

Step 5.3

Combine the numerators over the common denominator.

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{b^{2} \cdot 4 - 9}{4}}$

Step 5.4

Rewrite $\frac{b^{2} \cdot 4 - 9}{4}$ in a factored form.

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Step 5.4.1

Rewrite $b^{2} \cdot 4$ as ${(b \cdot 2)}^{2}$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{{(b \cdot 2)}^{2} - 9}{4}}$

Step 5.4.2

Rewrite $9$ as $3^{2}$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{{(b \cdot 2)}^{2} - 3^{2}}{4}}$

Step 5.4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b \cdot 2$ and $b = 3$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{(b \cdot 2 + 3) (b \cdot 2 - 3)}{4}}$

Step 5.4.4

Simplify.

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Step 5.4.4.1

Move $2$ to the left of $b$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{(2 \cdot b + 3) (b \cdot 2 - 3)}{4}}$

Step 5.4.4.2

Move $2$ to the left of $b$ .

$\frac{b - 3}{b + 2} \cdot \frac{1}{\frac{(2 b + 3) (2 b - 3)}{4}}$

Step 6

Multiply the numerator by the reciprocal of the denominator.

$\frac{b - 3}{b + 2} (1 \frac{4}{(2 b + 3) (2 b - 3)})$

Step 7

Multiply $\frac{4}{(2 b + 3) (2 b - 3)}$ by $1$ .

$\frac{b - 3}{b + 2} \cdot \frac{4}{(2 b + 3) (2 b - 3)}$

Step 8

Multiply $\frac{b - 3}{b + 2}$ by $\frac{4}{(2 b + 3) (2 b - 3)}$ .

$\frac{(b - 3) \cdot 4}{(b + 2) ((2 b + 3) (2 b - 3))}$

Step 9

Move $4$ to the left of $b - 3$ .

$\frac{4 (b - 3)}{(b + 2) (2 b + 3) (2 b - 3)}$