Calculus Examples

$\frac{B}{2 x^{3} + 14 x^{2}} \div \frac{5 x - 35}{10 x^{2} - 490} = 1$

Step 1

Multiply both sides by $\frac{5 x - 35}{10 x^{2} - 490}$ .

$\frac{B}{2 x^{3} + 14 x^{2}} \div \frac{5 x - 35}{10 x^{2} - 490} \cdot \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $\frac{B}{2 x^{3} + 14 x^{2}} \div \frac{5 x - 35}{10 x^{2} - 490} \cdot \frac{5 x - 35}{10 x^{2} - 490}$ .

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Step 2.1.1.1

To divide by a fraction, multiply by its reciprocal.

$\frac{B}{2 x^{3} + 14 x^{2}} \cdot \frac{10 x^{2} - 490}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2

Simplify terms.

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Step 2.1.1.2.1

Factor $2 x^{2}$ out of $2 x^{3} + 14 x^{2}$ .

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Step 2.1.1.2.1.1

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

$\frac{B}{2 x^{2} (x) + 14 x^{2}} \cdot \frac{10 x^{2} - 490}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.1.2

Factor $2 x^{2}$ out of $14 x^{2}$ .

$\frac{B}{2 x^{2} (x) + 2 x^{2} (7)} \cdot \frac{10 x^{2} - 490}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.1.3

Factor $2 x^{2}$ out of $2 x^{2} (x) + 2 x^{2} (7)$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{10 x^{2} - 490}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2

Cancel the common factor of $10 x^{2} - 490$ and $5 x - 35$ .

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Step 2.1.1.2.2.1

Factor $5$ out of $10 x^{2}$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2}) - 490}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.2

Factor $5$ out of $- 490$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2}) + 5 \cdot - 98}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.3

Factor $5$ out of $5 (2 x^{2}) + 5 (- 98)$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2} - 98)}{5 x - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.4

Cancel the common factors.

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Step 2.1.1.2.2.4.1

Factor $5$ out of $5 x$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2} - 98)}{5 (x) - 35} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.4.2

Factor $5$ out of $- 35$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2} - 98)}{5 x + 5 \cdot - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.4.3

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

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2} - 98)}{5 (x - 7)} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.4.4

Cancel the common factor.

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{5 (2 x^{2} - 98)}{5 (x - 7)} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.2.2.4.5

Rewrite the expression.

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{2 x^{2} - 98}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.3

Simplify the numerator.

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Step 2.1.1.3.1

Factor $2$ out of $2 x^{2} - 98$ .

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Step 2.1.1.3.1.1

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

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{2 (x^{2}) - 98}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.3.1.2

Factor $2$ out of $- 98$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{2 x^{2} + 2 \cdot - 49}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.3.1.3

Factor $2$ out of $2 x^{2} + 2 \cdot - 49$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{2 (x^{2} - 49)}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.3.2

Rewrite $49$ as $7^{2}$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{2 (x^{2} - 7^{2})}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.3.3

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 $b = 7$ .

$\frac{B}{2 x^{2} (x + 7)} \cdot \frac{2 (x + 7) (x - 7)}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4

Simplify terms.

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Step 2.1.1.4.1

Cancel the common factor of $2 (x + 7)$ .

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Step 2.1.1.4.1.1

Factor $2 (x + 7)$ out of $2 x^{2} (x + 7)$ .

$\frac{B}{2 (x + 7) (x^{2})} \cdot \frac{2 (x + 7) (x - 7)}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.1.2

Cancel the common factor.

$\frac{B}{2 (x + 7) x^{2}} \cdot \frac{2 (x + 7) (x - 7)}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.1.3

Rewrite the expression.

$\frac{B}{x^{2}} \cdot \frac{x - 7}{x - 7} \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.2

Multiply $\frac{B}{x^{2}}$ by $\frac{x - 7}{x - 7}$ .

$\frac{B (x - 7)}{x^{2} (x - 7)} \cdot \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.3

Cancel the common factor of $x - 7$ .

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Step 2.1.1.4.3.1

Cancel the common factor.

$\frac{B (x - 7)}{x^{2} (x - 7)} \cdot \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.3.2

Rewrite the expression.

$\frac{B}{x^{2}} \cdot \frac{5 x - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4

Cancel the common factor of $5 x - 35$ and $10 x^{2} - 490$ .

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Step 2.1.1.4.4.1

Factor $5$ out of $5 x$ .

$\frac{B}{x^{2}} \cdot \frac{5 (x) - 35}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.2

Factor $5$ out of $- 35$ .

$\frac{B}{x^{2}} \cdot \frac{5 x + 5 \cdot - 7}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.3

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

$\frac{B}{x^{2}} \cdot \frac{5 (x - 7)}{10 x^{2} - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.4

Cancel the common factors.

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Step 2.1.1.4.4.4.1

Factor $5$ out of $10 x^{2}$ .

$\frac{B}{x^{2}} \cdot \frac{5 (x - 7)}{5 (2 x^{2}) - 490} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.4.2

Factor $5$ out of $- 490$ .

$\frac{B}{x^{2}} \cdot \frac{5 (x - 7)}{5 (2 x^{2}) + 5 (- 98)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.4.3

Factor $5$ out of $5 (2 x^{2}) + 5 (- 98)$ .

$\frac{B}{x^{2}} \cdot \frac{5 (x - 7)}{5 (2 x^{2} - 98)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.4.4

Cancel the common factor.

$\frac{B}{x^{2}} \cdot \frac{5 (x - 7)}{5 (2 x^{2} - 98)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.4.4.4.5

Rewrite the expression.

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 x^{2} - 98} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.5

Simplify the denominator.

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Step 2.1.1.5.1

Factor $2$ out of $2 x^{2} - 98$ .

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Step 2.1.1.5.1.1

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

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 (x^{2}) - 98} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.5.1.2

Factor $2$ out of $- 98$ .

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 x^{2} + 2 \cdot - 49} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.5.1.3

Factor $2$ out of $2 x^{2} + 2 \cdot - 49$ .

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 (x^{2} - 49)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.5.2

Rewrite $49$ as $7^{2}$ .

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 (x^{2} - 7^{2})} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.5.3

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 $b = 7$ .

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 (x + 7) (x - 7)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.6

Simplify terms.

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Step 2.1.1.6.1

Cancel the common factor of $x - 7$ .

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Step 2.1.1.6.1.1

Cancel the common factor.

$\frac{B}{x^{2}} \cdot \frac{x - 7}{2 (x + 7) (x - 7)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.6.1.2

Rewrite the expression.

$\frac{B}{x^{2}} \cdot \frac{1}{2 (x + 7)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.6.2

Combine.

$\frac{B \cdot 1}{x^{2} (2 (x + 7))} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.6.3

Simplify the expression.

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Step 2.1.1.6.3.1

Multiply $B$ by $1$ .

$\frac{B}{x^{2} (2 (x + 7))} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.1.1.6.3.2

Move $2$ to the left of $x^{2}$ .

$\frac{B}{2 x^{2} (x + 7)} = 1 \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $1 \frac{5 x - 35}{10 x^{2} - 490}$ .

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Step 2.2.1.1

Reduce the expression by cancelling the common factors.

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Step 2.2.1.1.1

Multiply $\frac{5 x - 35}{10 x^{2} - 490}$ by $1$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 x - 35}{10 x^{2} - 490}$

Step 2.2.1.1.2

Cancel the common factor of $5 x - 35$ and $10 x^{2} - 490$ .

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Step 2.2.1.1.2.1

Factor $5$ out of $5 x$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 (x) - 35}{10 x^{2} - 490}$

Step 2.2.1.1.2.2

Factor $5$ out of $- 35$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 x + 5 \cdot - 7}{10 x^{2} - 490}$

Step 2.2.1.1.2.3

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

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 (x - 7)}{10 x^{2} - 490}$

Step 2.2.1.1.2.4

Cancel the common factors.

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Step 2.2.1.1.2.4.1

Factor $5$ out of $10 x^{2}$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 (x - 7)}{5 (2 x^{2}) - 490}$

Step 2.2.1.1.2.4.2

Factor $5$ out of $- 490$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 (x - 7)}{5 (2 x^{2}) + 5 (- 98)}$

Step 2.2.1.1.2.4.3

Factor $5$ out of $5 (2 x^{2}) + 5 (- 98)$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 (x - 7)}{5 (2 x^{2} - 98)}$

Step 2.2.1.1.2.4.4

Cancel the common factor.

$\frac{B}{2 x^{2} (x + 7)} = \frac{5 (x - 7)}{5 (2 x^{2} - 98)}$

Step 2.2.1.1.2.4.5

Rewrite the expression.

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 x^{2} - 98}$

Step 2.2.1.2

Simplify the denominator.

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Step 2.2.1.2.1

Factor $2$ out of $2 x^{2} - 98$ .

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Step 2.2.1.2.1.1

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

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 (x^{2}) - 98}$

Step 2.2.1.2.1.2

Factor $2$ out of $- 98$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 x^{2} + 2 \cdot - 49}$

Step 2.2.1.2.1.3

Factor $2$ out of $2 x^{2} + 2 \cdot - 49$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 (x^{2} - 49)}$

Step 2.2.1.2.2

Rewrite $49$ as $7^{2}$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 (x^{2} - 7^{2})}$

Step 2.2.1.2.3

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 $b = 7$ .

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 (x + 7) (x - 7)}$

Step 2.2.1.3

Cancel the common factor of $x - 7$ .

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Step 2.2.1.3.1

Cancel the common factor.

$\frac{B}{2 x^{2} (x + 7)} = \frac{x - 7}{2 (x + 7) (x - 7)}$

Step 2.2.1.3.2

Rewrite the expression.

$\frac{B}{2 x^{2} (x + 7)} = \frac{1}{2 (x + 7)}$

Step 3

Solve for $B$ .

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Step 3.1

Multiply both sides by $2 x^{2} (x + 7)$ .

$\frac{B}{2 x^{2} (x + 7)} (2 x^{2} (x + 7)) = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2

Simplify.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $\frac{B}{2 x^{2} (x + 7)} (2 x^{2} (x + 7))$ .

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Step 3.2.1.1.1

Rewrite using the commutative property of multiplication.

$2 \frac{B}{2 x^{2} (x + 7)} (x^{2} (x + 7)) = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Factor $2$ out of $2 x^{2} (x + 7)$ .

$2 \frac{B}{2 (x^{2} (x + 7))} (x^{2} (x + 7)) = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2.1.1.2.2

Cancel the common factor.

$2 \frac{B}{2 (x^{2} (x + 7))} (x^{2} (x + 7)) = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2.1.1.2.3

Rewrite the expression.

$\frac{B}{x^{2} (x + 7)} (x^{2} (x + 7)) = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2.1.1.3

Cancel the common factor of $x^{2} (x + 7)$ .

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Step 3.2.1.1.3.1

Cancel the common factor.

$\frac{B}{x^{2} (x + 7)} (x^{2} (x + 7)) = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2.1.1.3.2

Rewrite the expression.

$B = \frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $\frac{1}{2 (x + 7)} (2 x^{2} (x + 7))$ .

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Step 3.2.2.1.1

Rewrite using the commutative property of multiplication.

$B = 2 \frac{1}{2 (x + 7)} (x^{2} (x + 7))$

Step 3.2.2.1.2

Cancel the common factor of $2$ .

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Step 3.2.2.1.2.1

Cancel the common factor.

$B = 2 \frac{1}{2 (x + 7)} (x^{2} (x + 7))$

Step 3.2.2.1.2.2

Rewrite the expression.

$B = \frac{1}{x + 7} (x^{2} (x + 7))$

Step 3.2.2.1.3

Cancel the common factor of $x + 7$ .

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Step 3.2.2.1.3.1

Factor $x + 7$ out of $x^{2} (x + 7)$ .

$B = \frac{1}{x + 7} ((x + 7) x^{2})$

Step 3.2.2.1.3.2

Cancel the common factor.

$B = \frac{1}{x + 7} ((x + 7) x^{2})$

Step 3.2.2.1.3.3

Rewrite the expression.

$B = x^{2}$