Precalculus Examples

$\frac{2 x^{2} - 5 x - 12}{3 x^{2} - 11 x - 4}$

Step 1

Set the denominator in $\frac{2 x^{2} - 5 x - 12}{3 x^{2} - 11 x - 4}$ equal to $0$ to find where the expression is undefined.

$3 x^{2} - 11 x - 4 = 0$

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 $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = - 11$ .

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Step 2.1.1.1

Factor $- 11$ out of $- 11 x$ .

$3 x^{2} - 11 x - 4 = 0$

Step 2.1.1.2

Rewrite $- 11$ as $1$ plus $- 12$

$3 x^{2} + (1 - 12) x - 4 = 0$

Step 2.1.1.3

Apply the distributive property.

$3 x^{2} + 1 x - 12 x - 4 = 0$

Step 2.1.1.4

Multiply $x$ by $1$ .

$3 x^{2} + x - 12 x - 4 = 0$

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.

$(3 x^{2} + x) - 12 x - 4 = 0$

Step 2.1.2.2

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

$x (3 x + 1) - 4 (3 x + 1) = 0$

Step 2.1.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$(3 x + 1) (x - 4) = 0$

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 $0$ .

$3 x + 1 = 0$

$x - 4 = 0$

Step 2.3

Set $3 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Step 2.3.2

Solve $3 x + 1 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Step 2.3.2.2

Divide each term in $3 x = - 1$ by $3$ and simplify.

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Step 2.3.2.2.1

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

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 $3$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

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

Add $4$ to both sides of the equation.

$x = 4$

Step 2.5

The final solution is all the values that make $(3 x + 1) (x - 4) = 0$ true.

$x = - \frac{1}{3}, 4$

Step 3

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

Interval Notation:

$(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x \neq - \frac{1}{3}, 4}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, \frac{11}{13}) \cup (\frac{11}{13}, \infty)$

Set-Builder Notation:

${y | y \neq \frac{2}{3}, \frac{11}{13}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 4) \cup (4, \infty), {x | x \neq - \frac{1}{3}, 4}$

Range: $(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, \frac{11}{13}) \cup (\frac{11}{13}, \infty), {y | y \neq \frac{2}{3}, \frac{11}{13}}$

Step 6