Precalculus Examples

$\frac{(x + 5) (x^{2} - 25)}{{(x + 5)}^{2} {(x - 5)}^{2}}$

Step 1

Set the denominator in $\frac{(x + 5) (x^{2} - 25)}{{(x + 5)}^{2} {(x - 5)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 5)}^{2} {(x - 5)}^{2} = 0$

Step 2

Solve for $x$ .

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Step 2.1

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

${(x + 5)}^{2} = 0$

${(x - 5)}^{2} = 0$

Step 2.2

Set ${(x + 5)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.2.1

Set ${(x + 5)}^{2}$ equal to $0$ .

${(x + 5)}^{2} = 0$

Step 2.2.2

Solve ${(x + 5)}^{2} = 0$ for $x$ .

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Step 2.2.2.1

Set the $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.2.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.3

Set ${(x - 5)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set ${(x - 5)}^{2}$ equal to $0$ .

${(x - 5)}^{2} = 0$

Step 2.3.2

Solve ${(x - 5)}^{2} = 0$ for $x$ .

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Step 2.3.2.1

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.3.2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.4

The final solution is all the values that make ${(x + 5)}^{2} {(x - 5)}^{2} = 0$ true.

$x = - 5, 5$

Step 3

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5, - 5}$

Step 4