Precalculus Examples

$\frac{b^{3} - 8}{b^{2} - 9} \cdot \frac{b + 3}{b^{2} + 2 b + 4}$

Step 1

Set the denominator in $\frac{b^{3} - 8}{b^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$b^{2} - 9 = 0$

Step 2

Solve for $b$ .

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Step 2.1

Add $9$ to both sides of the equation.

$b^{2} = 9$

Step 2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{9}$

Step 2.3

Simplify $\pm \sqrt{9}$ .

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Step 2.3.1

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

$b = \pm \sqrt{3^{2}}$

Step 2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm 3$

Step 2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$b = 3$

Step 2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$b = - 3$

Step 2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$b = 3, - 3$

Step 3

Set the denominator in $\frac{b + 3}{b^{2} + 2 b + 4}$ equal to $0$ to find where the expression is undefined.

$b^{2} + 2 b + 4 = 0$

Step 4

Solve for $b$ .

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Step 4.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.2

Substitute the values $a = 1$ , $b = 2$ , and $c = 4$ into the quadratic formula and solve for $b$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Step 4.3

Simplify.

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Step 4.3.1

Simplify the numerator.

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Step 4.3.1.1

Raise $2$ to the power of $2$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 4.3.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 4.3.1.2.1

Multiply $- 4$ by $1$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 4.3.1.2.2

Multiply $- 4$ by $4$ .

$b = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 4.3.1.3

Subtract $16$ from $4$ .

$b = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 4.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$b = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 4.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$b = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$b = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.3.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 4.3.1.7.1

Factor $4$ out of $12$ .

$b = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 4.3.1.7.2

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

$b = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 4.3.1.8

Pull terms out from under the radical.

$b = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 4.3.1.9

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

$b = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 4.3.2

Multiply $2$ by $1$ .

$b = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 4.3.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$b = - 1 \pm i \sqrt{3}$

Step 4.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 4.4.1

Simplify the numerator.

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Step 4.4.1.1

Raise $2$ to the power of $2$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 4.4.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 4.4.1.2.1

Multiply $- 4$ by $1$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 4.4.1.2.2

Multiply $- 4$ by $4$ .

$b = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 4.4.1.3

Subtract $16$ from $4$ .

$b = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 4.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$b = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 4.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$b = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$b = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 4.4.1.7.1

Factor $4$ out of $12$ .

$b = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 4.4.1.7.2

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

$b = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 4.4.1.8

Pull terms out from under the radical.

$b = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 4.4.1.9

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

$b = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 4.4.2

Multiply $2$ by $1$ .

$b = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 4.4.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$b = - 1 \pm i \sqrt{3}$

Step 4.4.4

Change the $\pm$ to $+$ .

$b = - 1 + i \sqrt{3}$

Step 4.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $2$ to the power of $2$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 4.5.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $1$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 4.5.1.2.2

Multiply $- 4$ by $4$ .

$b = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 4.5.1.3

Subtract $16$ from $4$ .

$b = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 4.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$b = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 4.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$b = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$b = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 4.5.1.7.1

Factor $4$ out of $12$ .

$b = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 4.5.1.7.2

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

$b = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 4.5.1.8

Pull terms out from under the radical.

$b = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 4.5.1.9

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

$b = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 4.5.2

Multiply $2$ by $1$ .

$b = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 4.5.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$b = - 1 \pm i \sqrt{3}$

Step 4.5.4

Change the $\pm$ to $-$ .

$b = - 1 - i \sqrt{3}$

Step 4.6

The final answer is the combination of both solutions.

$b = - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 5

The domain is all values of $b$ that make the expression defined.

Interval Notation:

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Set-Builder Notation:

${b | b \neq 3, - 3}$

Step 6