Finite Math Examples

$\frac{6 - 3 x}{6 x + 6 x^{2}} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 1

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

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Step 1.1

Factor $3$ out of $6$ .

$\frac{3 (2) - 3 x}{6 x + 6 x^{2}} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 1.2

Factor $3$ out of $- 3 x$ .

$\frac{3 (2) + 3 (- x)}{6 x + 6 x^{2}} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 1.3

Factor $3$ out of $3 (2) + 3 (- x)$ .

$\frac{3 (2 - x)}{6 x + 6 x^{2}} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 2

Factor $6 x$ out of $6 x + 6 x^{2}$ .

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Step 2.1

Factor $6 x$ out of $6 x$ .

$\frac{3 (2 - x)}{6 x (1) + 6 x^{2}} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 2.2

Factor $6 x$ out of $6 x^{2}$ .

$\frac{3 (2 - x)}{6 x (1) + 6 x (x)} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 2.3

Factor $6 x$ out of $6 x (1) + 6 x (x)$ .

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{3 x^{2} - 2 x - 5}{x^{2} - 4}$

Step 3

Factor by grouping.

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Step 3.1

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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = - 2$ .

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Step 3.1.1

Factor $- 2$ out of $- 2 x$ .

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{3 x^{2} - 2 (x) - 5}{x^{2} - 4}$

Step 3.1.2

Rewrite $- 2$ as $3$ plus $- 5$

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{3 x^{2} + (3 - 5) x - 5}{x^{2} - 4}$

Step 3.1.3

Apply the distributive property.

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{3 x^{2} + 3 x - 5 x - 5}{x^{2} - 4}$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{(3 x^{2} + 3 x) - 5 x - 5}{x^{2} - 4}$

Step 3.2.2

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

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{3 x (x + 1) - 5 (x + 1)}{x^{2} - 4}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{(x + 1) (3 x - 5)}{x^{2} - 4}$

Step 4

Simplify the denominator.

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Step 4.1

Rewrite $4$ as $2^{2}$ .

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{(x + 1) (3 x - 5)}{x^{2} - 2^{2}}$

Step 4.2

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 $b = 2$ .

$\frac{3 (2 - x)}{6 x (1 + x)} \cdot \frac{(x + 1) (3 x - 5)}{(x + 2) (x - 2)}$

Step 5

Cancel the common factors.

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Step 5.1

Factor $3$ out of $6 x (1 + x)$ .

$\frac{3 (2 - x)}{3 (2 x (1 + x))} \cdot \frac{(x + 1) (3 x - 5)}{(x + 2) (x - 2)}$

Step 5.2

Cancel the common factor.

$\frac{3 (2 - x)}{3 (2 x (1 + x))} \cdot \frac{(x + 1) (3 x - 5)}{(x + 2) (x - 2)}$

Step 5.3

Rewrite the expression.

$\frac{2 - x}{2 x (1 + x)} \cdot \frac{(x + 1) (3 x - 5)}{(x + 2) (x - 2)}$

Step 6

Multiply $\frac{2 - x}{2 x (1 + x)}$ by $\frac{(x + 1) (3 x - 5)}{(x + 2) (x - 2)}$ .

$\frac{(2 - x) ((x + 1) (3 x - 5))}{2 x (1 + x) ((x + 2) (x - 2))}$

Step 7

Cancel the common factor of $2 - x$ and $x - 2$ .

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Step 7.1

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{(- 1 (- 2) - x) ((x + 1) (3 x - 5))}{2 x (1 + x) ((x + 2) (x - 2))}$

Step 7.2

Factor $- 1$ out of $- x$ .

$\frac{(- 1 (- 2) - (x)) ((x + 1) (3 x - 5))}{2 x (1 + x) ((x + 2) (x - 2))}$

Step 7.3

Factor $- 1$ out of $- 1 (- 2) - (x)$ .

$\frac{- 1 (- 2 + x) ((x + 1) (3 x - 5))}{2 x (1 + x) ((x + 2) (x - 2))}$

Step 7.4

Reorder terms.

$\frac{- 1 (x - 2) ((x + 1) (3 x - 5))}{2 x (1 + x) ((x + 2) (x - 2))}$

Step 7.5

Cancel the common factor.

$\frac{- 1 (x - 2) ((x + 1) (3 x - 5))}{2 x (1 + x) ((x + 2) (x - 2))}$

Step 7.6

Rewrite the expression.

$\frac{- 1 ((x + 1) (3 x - 5))}{2 x (1 + x) (x + 2)}$

Step 8

Cancel the common factor of $x + 1$ and $1 + x$ .

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Step 8.1

Reorder terms.

$\frac{- 1 ((x + 1) (3 x - 5))}{2 x (x + 1) (x + 2)}$

Step 8.2

Cancel the common factor.

$\frac{- 1 ((x + 1) (3 x - 5))}{2 x (x + 1) (x + 2)}$

Step 8.3

Rewrite the expression.

$\frac{- 1 (3 x - 5)}{2 x (x + 2)}$

Step 9

Move the negative in front of the fraction.

$- \frac{3 x - 5}{2 x (x + 2)}$