Precalculus Examples

$f (x) = \frac{x^{2} + x - 6}{x^{3} - 3 x^{2} - 16 x + 48}$

Step 1

Set the denominator in $\frac{x^{2} + x - 6}{x^{3} - 3 x^{2} - 16 x + 48}$ equal to $0$ to find where the expression is undefined.

$x^{3} - 3 x^{2} - 16 x + 48 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor out the greatest common factor from each group.

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Step 2.1.1.1

Group the first two terms and the last two terms.

$(x^{3} - 3 x^{2}) - 16 x + 48 = 0$

Step 2.1.1.2

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 3) - 16 (x - 3) = 0$

Step 2.1.2

Factor the polynomial by factoring out the greatest common factor, $x - 3$ .

$(x - 3) (x^{2} - 16) = 0$

Step 2.1.3

Rewrite $16$ as $4^{2}$ .

$(x - 3) (x^{2} - 4^{2}) = 0$

Step 2.1.4

Factor.

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Step 2.1.4.1

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 $b = 4$ .

$(x - 3) ((x + 4) (x - 4)) = 0$

Step 2.1.4.2

Remove unnecessary parentheses.

$(x - 3) (x + 4) (x - 4) = 0$

Step 2.2

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

$x - 3 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 2.3

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

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Step 2.3.1

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

$x - 3 = 0$

Step 2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.4

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

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Step 2.4.1

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

$x + 4 = 0$

Step 2.4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.5

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

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Step 2.5.1

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

$x - 4 = 0$

Step 2.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.6

The final solution is all the values that make $(x - 3) (x + 4) (x - 4) = 0$ true.

$x = 3, - 4, 4$

Step 3

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, 3) \cup (3, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x \neq 3, 4, - 4}$

Step 4