Precalculus Examples

$f (x) = \frac{2 x^{2} - 3}{x^{3} + 3 x^{2} + 3 x + 1}$

Step 1

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

$x^{3} + 3 x^{2} + 3 x + 1 = 0$

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

Regroup terms.

$x^{3} + 1 + 3 x^{2} + 3 x = 0$

Step 2.1.2

Rewrite $1$ as $1^{3}$ .

$x^{3} + 1^{3} + 3 x^{2} + 3 x = 0$

Step 2.1.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

$(x + 1) (x^{2} - x \cdot 1 + 1^{2}) + 3 x^{2} + 3 x = 0$

Step 2.1.4

Simplify.

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Step 2.1.4.1

Multiply $- 1$ by $1$ .

$(x + 1) (x^{2} - x + 1^{2}) + 3 x^{2} + 3 x = 0$

Step 2.1.4.2

One to any power is one.

$(x + 1) (x^{2} - x + 1) + 3 x^{2} + 3 x = 0$

Step 2.1.5

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

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Step 2.1.5.1

Factor $3 x$ out of $3 x^{2}$ .

$(x + 1) (x^{2} - x + 1) + 3 x (x) + 3 x = 0$

Step 2.1.5.2

Factor $3 x$ out of $3 x$ .

$(x + 1) (x^{2} - x + 1) + 3 x (x) + 3 x (1) = 0$

Step 2.1.5.3

Factor $3 x$ out of $3 x (x) + 3 x (1)$ .

$(x + 1) (x^{2} - x + 1) + 3 x (x + 1) = 0$

Step 2.1.6

Factor $x + 1$ out of $(x + 1) (x^{2} - x + 1) + 3 x (x + 1)$ .

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Step 2.1.6.1

Factor $x + 1$ out of $3 x (x + 1)$ .

$(x + 1) (x^{2} - x + 1) + (x + 1) (3 x) = 0$

Step 2.1.6.2

Factor $x + 1$ out of $(x + 1) (x^{2} - x + 1) + (x + 1) (3 x)$ .

$(x + 1) (x^{2} - x + 1 + 3 x) = 0$

Step 2.1.7

Add $- x$ and $3 x$ .

$(x + 1) (x^{2} + 2 x + 1) = 0$

Step 2.1.8

Factor using the perfect square rule.

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Step 2.1.8.1

Rewrite $1$ as $1^{2}$ .

$(x + 1) (x^{2} + 2 x + 1^{2}) = 0$

Step 2.1.8.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 2.1.8.3

Rewrite the polynomial.

$(x + 1) (x^{2} + 2 \cdot x \cdot 1 + 1^{2}) = 0$

Step 2.1.8.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

$(x + 1) {(x + 1)}^{2} = 0$

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 $0$ .

$x + 1 = 0$

${(x + 1)}^{2} = 0$

Step 2.3

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4

The final solution is all the values that make $(x + 1) {(x + 1)}^{2} = 0$ true.

$x = - 1$

Step 3

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, \infty)$

Set-Builder Notation:

${x | x \neq - 1}$

Step 4