Precalculus Examples

$\frac{x^{2} - 1}{x^{3} + 2 x^{2} - 8 x} - \frac{\sqrt{x + 1}}{1 - x}$

Step 1

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Step 2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 3

Set the denominator in $\frac{x^{2} - 1}{x^{3} + 2 x^{2} - 8 x}$ equal to $0$ to find where the expression is undefined.

$x^{3} + 2 x^{2} - 8 x = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor the left side of the equation.

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Step 4.1.1

Factor $x$ out of $x^{3} + 2 x^{2} - 8 x$ .

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Step 4.1.1.1

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

$x \cdot x^{2} + 2 x^{2} - 8 x = 0$

Step 4.1.1.2

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

$x \cdot x^{2} + x (2 x) - 8 x = 0$

Step 4.1.1.3

Factor $x$ out of $- 8 x$ .

$x \cdot x^{2} + x (2 x) + x \cdot - 8 = 0$

Step 4.1.1.4

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

$x (x^{2} + 2 x) + x \cdot - 8 = 0$

Step 4.1.1.5

Factor $x$ out of $x (x^{2} + 2 x) + x \cdot - 8$ .

$x (x^{2} + 2 x - 8) = 0$

Step 4.1.2

Factor.

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Step 4.1.2.1

Factor $x^{2} + 2 x - 8$ using the AC method.

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Step 4.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

Step 4.1.2.1.2

Write the factored form using these integers.

$x ((x - 2) (x + 4)) = 0$

Step 4.1.2.2

Remove unnecessary parentheses.

$x (x - 2) (x + 4) = 0$

Step 4.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0$

$x + 4 = 0$

Step 4.3

Set $x$ equal to $0$ .

$x = 0$

Step 4.4

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 4.4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 4.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 4.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 4.6

The final solution is all the values that make $x (x - 2) (x + 4) = 0$ true.

$x = 0, 2, - 4$

Step 5

Set the denominator in $\frac{\sqrt{x + 1}}{1 - x}$ equal to $0$ to find where the expression is undefined.

$1 - x = 0$

Step 6

Solve for $x$ .

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Step 6.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 6.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 6.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 6.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 7

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

Interval Notation:

$[- 1, 0) \cup (0, 1) \cup (1, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \geq - 1, x \neq 0, 1, 2}$

Step 8