Precalculus Examples

$f (x) = \frac{x^{2} - 3 x - 4}{2 x^{2} + x - 1}$

Step 1

Set the denominator in $\frac{x^{2} - 3 x - 4}{2 x^{2} + x - 1}$ equal to $0$ to find where the expression is undefined.

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

Step 2

Solve for $x$ .

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Step 2.1

Factor by grouping.

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Step 2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Step 2.1.1.1

Multiply by $1$ .

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

Step 2.1.1.2

Rewrite $1$ as $- 1$ plus $2$

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

Step 2.1.1.3

Apply the distributive property.

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

Step 2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1

Group the first two terms and the last two terms.

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

Step 2.1.2.2

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

$x (2 x - 1) + 1 (2 x - 1) = 0$

Step 2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x + 1) = 0$

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 $0$ .

$2 x - 1 = 0$

$x + 1 = 0$

Step 2.3

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 2.3.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 2.3.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.3.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 2.3.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 2.4

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

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Step 2.4.1

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

$x + 1 = 0$

Step 2.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5

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

$x = \frac{1}{2}, - 1$

Step 3

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Set-Builder Notation:

${x | x \neq \frac{1}{2}, - 1}$

Step 4