Finite Math Examples

$\frac{9 x^{2} + 58 x - 35}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \div \frac{5 x + 35}{3 x^{3}}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{9 x^{2} + 58 x - 35}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 2

Factor by grouping.

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Step 2.1

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 $a \cdot c = 9 \cdot - 35 = - 315$ and whose sum is $b = 58$ .

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Step 2.1.1

Factor $58$ out of $58 x$ .

$\frac{9 x^{2} + 58 (x) - 35}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 2.1.2

Rewrite $58$ as $- 5$ plus $63$

$\frac{9 x^{2} + (- 5 + 63) x - 35}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 2.1.3

Apply the distributive property.

$\frac{9 x^{2} - 5 x + 63 x - 35}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

$\frac{(9 x^{2} - 5 x) + 63 x - 35}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 2.2.2

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

$\frac{x (9 x - 5) + 7 (9 x - 5)}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $9 x - 5$ .

$\frac{(9 x - 5) (x + 7)}{7 x - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 3

Simplify with factoring out.

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Step 3.1

Factor $7$ out of $7 x - 35$ .

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Step 3.1.1

Factor $7$ out of $7 x$ .

$\frac{(9 x - 5) (x + 7)}{7 (x) - 35} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 3.1.2

Factor $7$ out of $- 35$ .

$\frac{(9 x - 5) (x + 7)}{7 x + 7 \cdot - 5} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 3.1.3

Factor $7$ out of $7 x + 7 \cdot - 5$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x^{2} - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 3.2

Factor $x$ out of $x^{2} - 5 x$ .

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Step 3.2.1

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

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x \cdot x - 5 x}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 3.2.2

Factor $x$ out of $- 5 x$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x \cdot x + x \cdot - 5}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 3.2.3

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x (x - 5)}{81 x^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 4

Simplify the denominator.

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Step 4.1

Rewrite $81 x^{2}$ as ${(9 x)}^{2}$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x (x - 5)}{{(9 x)}^{2} - 25} \frac{3 x^{3}}{5 x + 35}$

Step 4.2

Rewrite $25$ as $5^{2}$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x (x - 5)}{{(9 x)}^{2} - 5^{2}} \frac{3 x^{3}}{5 x + 35}$

Step 4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 9 x$ and $b = 5$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x (x - 5)}{(9 x + 5) (9 x - 5)} \frac{3 x^{3}}{5 x + 35}$

Step 5

Simplify terms.

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Step 5.1

Cancel the common factor of $9 x - 5$ .

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Step 5.1.1

Factor $9 x - 5$ out of $(9 x + 5) (9 x - 5)$ .

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x (x - 5)}{(9 x - 5) (9 x + 5)} \frac{3 x^{3}}{5 x + 35}$

Step 5.1.2

Cancel the common factor.

$\frac{(9 x - 5) (x + 7)}{7 (x - 5)} \cdot \frac{x (x - 5)}{(9 x - 5) (9 x + 5)} \frac{3 x^{3}}{5 x + 35}$

Step 5.1.3

Rewrite the expression.

$\frac{x + 7}{7 (x - 5)} \cdot \frac{x (x - 5)}{9 x + 5} \frac{3 x^{3}}{5 x + 35}$

Step 5.2

Cancel the common factor of $x - 5$ .

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Step 5.2.1

Factor $x - 5$ out of $7 (x - 5)$ .

$\frac{x + 7}{(x - 5) \cdot 7} \cdot \frac{x (x - 5)}{9 x + 5} \frac{3 x^{3}}{5 x + 35}$

Step 5.2.2

Factor $x - 5$ out of $x (x - 5)$ .

$\frac{x + 7}{(x - 5) \cdot 7} \cdot \frac{(x - 5) x}{9 x + 5} \frac{3 x^{3}}{5 x + 35}$

Step 5.2.3

Cancel the common factor.

$\frac{x + 7}{(x - 5) \cdot 7} \cdot \frac{(x - 5) x}{9 x + 5} \frac{3 x^{3}}{5 x + 35}$

Step 5.2.4

Rewrite the expression.

$\frac{x + 7}{7} \cdot \frac{x}{9 x + 5} \frac{3 x^{3}}{5 x + 35}$

Step 5.3

Multiply $\frac{x + 7}{7}$ by $\frac{x}{9 x + 5}$ .

$\frac{(x + 7) x}{7 (9 x + 5)} \cdot \frac{3 x^{3}}{5 x + 35}$

Step 5.4

Factor $5$ out of $5 x + 35$ .

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Step 5.4.1

Factor $5$ out of $5 x$ .

$\frac{(x + 7) x}{7 (9 x + 5)} \cdot \frac{3 x^{3}}{5 (x) + 35}$

Step 5.4.2

Factor $5$ out of $35$ .

$\frac{(x + 7) x}{7 (9 x + 5)} \cdot \frac{3 x^{3}}{5 x + 5 \cdot 7}$

Step 5.4.3

Factor $5$ out of $5 x + 5 \cdot 7$ .

$\frac{(x + 7) x}{7 (9 x + 5)} \cdot \frac{3 x^{3}}{5 (x + 7)}$

Step 5.5

Combine.

$\frac{(x + 7) x (3 x^{3})}{7 (9 x + 5) (5 (x + 7))}$

Step 6

Multiply $x$ by $x^{3}$ by adding the exponents.

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Step 6.1

Move $x^{3}$ .

$\frac{(x + 7) (x^{3} x) \cdot 3}{7 (9 x + 5) (5 (x + 7))}$

Step 6.2

Multiply $x^{3}$ by $x$ .

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Step 6.2.1

Raise $x$ to the power of $1$ .

$\frac{(x + 7) (x^{3} x^{1}) \cdot 3}{7 (9 x + 5) (5 (x + 7))}$

Step 6.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{(x + 7) x^{3 + 1} \cdot 3}{7 (9 x + 5) (5 (x + 7))}$

Step 6.3

Add $3$ and $1$ .

$\frac{(x + 7) x^{4} \cdot 3}{7 (9 x + 5) (5 (x + 7))}$

Step 7

Cancel the common factor of $x + 7$ .

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Step 7.1

Cancel the common factor.

$\frac{(x + 7) x^{4} \cdot 3}{7 (9 x + 5) (5 (x + 7))}$

Step 7.2

Rewrite the expression.

$\frac{x^{4} \cdot 3}{7 (9 x + 5) \cdot (5)}$

Step 8

Simplify the expression.

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Step 8.1

Multiply $5$ by $7$ .

$\frac{x^{4} \cdot 3}{35 (9 x + 5)}$

Step 8.2

Move $3$ to the left of $x^{4}$ .

$\frac{3 x^{4}}{35 (9 x + 5)}$