Finite Math Examples

$(\frac{a^{2} - 7 a b - 18 b^{2}}{- a + 9 b}) \cdot (\frac{3 a - 6 b}{a^{2} - 4 b^{2}})$

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 1 \cdot - 18 = - 18$ and whose sum is $b = - 7$ .

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Step 1.1.1

Reorder terms.

$\frac{a^{2} - 18 b^{2} - 7 a b}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.1.2

Reorder $- 18 b^{2}$ and $- 7 a b$ .

$\frac{a^{2} - 7 a b - 18 b^{2}}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.1.3

Factor $- 7$ out of $- 7 a b$ .

$\frac{a^{2} - 7 (a b) - 18 b^{2}}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.1.4

Rewrite $- 7$ as $2$ plus $- 9$

$\frac{a^{2} + (2 - 9) (a b) - 18 b^{2}}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.1.5

Apply the distributive property.

$\frac{a^{2} + 2 (a b) - 9 (a b) - 18 b^{2}}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.1.6

Move parentheses.

$\frac{a^{2} + 2 a b - 9 a b - 18 b^{2}}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$\frac{(a^{2} + 2 a b) - 9 a b - 18 b^{2}}{- a + 9 b} \cdot \frac{3 a - 6 b}{a^{2} - 4 b^{2}}$

Step 1.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $a + 2 b$ .

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

Step 2

Factor $3$ out of $3 a - 6 b$ .

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Step 2.1

Factor $3$ out of $3 a$ .

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

Step 2.2

Factor $3$ out of $- 6 b$ .

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

Step 2.3

Factor $3$ out of $3 (a) + 3 (- 2 b)$ .

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

Step 3

Simplify the denominator.

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Step 3.1

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

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

Step 3.2

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

$\frac{(a + 2 b) (a - 9 b)}{- a + 9 b} \cdot \frac{3 (a - 2 b)}{(a + 2 b) (a - (2 b))}$

Step 3.3

Multiply $2$ by $- 1$ .

$\frac{(a + 2 b) (a - 9 b)}{- a + 9 b} \cdot \frac{3 (a - 2 b)}{(a + 2 b) (a - 2 b)}$

Step 4

Simplify terms.

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Step 4.1

Cancel the common factor of $a + 2 b$ .

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Step 4.1.1

Cancel the common factor.

$\frac{(a + 2 b) (a - 9 b)}{- a + 9 b} \cdot \frac{3 (a - 2 b)}{(a + 2 b) (a - 2 b)}$

Step 4.1.2

Rewrite the expression.

$\frac{a - 9 b}{- a + 9 b} \cdot \frac{3 (a - 2 b)}{a - 2 b}$

Step 4.2

Multiply $\frac{a - 9 b}{- a + 9 b}$ by $\frac{3 (a - 2 b)}{a - 2 b}$ .

$\frac{(a - 9 b) (3 (a - 2 b))}{(- a + 9 b) (a - 2 b)}$

Step 4.3

Cancel the common factor of $a - 2 b$ .

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Step 4.3.1

Cancel the common factor.

$\frac{(a - 9 b) (3 (a - 2 b))}{(- a + 9 b) (a - 2 b)}$

Step 4.3.2

Rewrite the expression.

$\frac{(a - 9 b) \cdot (3)}{- a + 9 b}$

Step 4.4

Cancel the common factor of $a - 9 b$ and $- a + 9 b$ .

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Step 4.4.1

Factor $- 1$ out of $a$ .

$\frac{(- 1 (- a) - 9 b) \cdot 3}{- a + 9 b}$

Step 4.4.2

Factor $- 1$ out of $- 9 b$ .

$\frac{(- 1 (- a) - (9 b)) \cdot 3}{- a + 9 b}$

Step 4.4.3

Factor $- 1$ out of $- 1 (- a) - (9 b)$ .

$\frac{- 1 (- a + 9 b) \cdot 3}{- a + 9 b}$

Step 4.4.4

Cancel the common factor.

$\frac{- 1 (- a + 9 b) \cdot 3}{- a + 9 b}$

Step 4.4.5

Divide $- 1 \cdot 3$ by $1$ .

$- 1 \cdot 3$

Step 4.5

Multiply $- 1$ by $3$ .

$- 3$