Precalculus Examples

$\frac{a^{2} - 25}{a + 3} \cdot \frac{1}{a^{2} + 5 a} - \frac{a + 5}{a^{2} - 3 a}$

Step 1

Set the denominator in $\frac{a^{2} - 25}{a + 3}$ equal to $0$ to find where the expression is undefined.

$a + 3 = 0$

Step 2

Subtract $3$ from both sides of the equation.

$a = - 3$

Step 3

Set the denominator in $\frac{1}{a^{2} + 5 a}$ equal to $0$ to find where the expression is undefined.

$a^{2} + 5 a = 0$

Step 4

Solve for $a$ .

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Step 4.1

Factor $a$ out of $a^{2} + 5 a$ .

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Step 4.1.1

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

$a \cdot a + 5 a = 0$

Step 4.1.2

Factor $a$ out of $5 a$ .

$a \cdot a + a \cdot 5 = 0$

Step 4.1.3

Factor $a$ out of $a \cdot a + a \cdot 5$ .

$a (a + 5) = 0$

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 $0$ .

$a = 0$

$a + 5 = 0$

Step 4.3

Set $a$ equal to $0$ .

$a = 0$

Step 4.4

Set $a + 5$ equal to $0$ and solve for $a$ .

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Step 4.4.1

Set $a + 5$ equal to $0$ .

$a + 5 = 0$

Step 4.4.2

Subtract $5$ from both sides of the equation.

$a = - 5$

Step 4.5

The final solution is all the values that make $a (a + 5) = 0$ true.

$a = 0, - 5$

Step 5

Set the denominator in $\frac{a + 5}{a^{2} - 3 a}$ equal to $0$ to find where the expression is undefined.

$a^{2} - 3 a = 0$

Step 6

Solve for $a$ .

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Step 6.1

Factor $a$ out of $a^{2} - 3 a$ .

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Step 6.1.1

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

$a \cdot a - 3 a = 0$

Step 6.1.2

Factor $a$ out of $- 3 a$ .

$a \cdot a + a \cdot - 3 = 0$

Step 6.1.3

Factor $a$ out of $a \cdot a + a \cdot - 3$ .

$a (a - 3) = 0$

Step 6.2

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

$a = 0$

$a - 3 = 0$

Step 6.3

Set $a$ equal to $0$ .

$a = 0$

Step 6.4

Set $a - 3$ equal to $0$ and solve for $a$ .

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Step 6.4.1

Set $a - 3$ equal to $0$ .

$a - 3 = 0$

Step 6.4.2

Add $3$ to both sides of the equation.

$a = 3$

Step 6.5

The final solution is all the values that make $a (a - 3) = 0$ true.

$a = 0, 3$

Step 7

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, - 3) \cup (- 3, 0) \cup (0, 3) \cup (3, \infty)$

Set-Builder Notation:

${a | a \neq 0, 3, - 3, - 5}$

Step 8