Calculus Examples

$\frac{8 x^{3} - 27}{2 x^{2} - x - 3} < \frac{3}{2}$

Step 1

Subtract $\frac{3}{2}$ from both sides of the inequality.

$\frac{8 x^{3} - 27}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2

Simplify $\frac{8 x^{3} - 27}{2 x^{2} - x - 3} - \frac{3}{2}$ .

Tap for more steps...

Step 2.1

Simplify each term.

Tap for more steps...

Step 2.1.1

Simplify the numerator.

Tap for more steps...

Step 2.1.1.1

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

$\frac{{(2 x)}^{3} - 27}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.1.2

Rewrite $27$ as $3^{3}$ .

$\frac{{(2 x)}^{3} - 3^{3}}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.1.3

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

$\frac{(2 x - 3) ({(2 x)}^{2} + 2 x \cdot 3 + 3^{2})}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.1.4

Simplify.

Tap for more steps...

Step 2.1.1.4.1

Apply the product rule to $2 x$ .

$\frac{(2 x - 3) (2^{2} x^{2} + 2 x \cdot 3 + 3^{2})}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.1.4.2

Raise $2$ to the power of $2$ .

$\frac{(2 x - 3) (4 x^{2} + 2 x \cdot 3 + 3^{2})}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.1.4.3

Multiply $3$ by $2$ .

$\frac{(2 x - 3) (4 x^{2} + 6 x + 3^{2})}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.1.4.4

Raise $3$ to the power of $2$ .

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{2 x^{2} - x - 3} - \frac{3}{2} < 0$

Step 2.1.2

Factor by grouping.

Tap for more steps...

Step 2.1.2.1

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 1$ .

Tap for more steps...

Step 2.1.2.1.1

Factor $- 1$ out of $- x$ .

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{2 x^{2} - (x) - 3} - \frac{3}{2} < 0$

Step 2.1.2.1.2

Rewrite $- 1$ as $2$ plus $- 3$

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{2 x^{2} + (2 - 3) x - 3} - \frac{3}{2} < 0$

Step 2.1.2.1.3

Apply the distributive property.

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{2 x^{2} + 2 x - 3 x - 3} - \frac{3}{2} < 0$

Step 2.1.2.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.2.2.1

Group the first two terms and the last two terms.

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{(2 x^{2} + 2 x) - 3 x - 3} - \frac{3}{2} < 0$

Step 2.1.2.2.2

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

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{2 x (x + 1) - 3 (x + 1)} - \frac{3}{2} < 0$

Step 2.1.2.3

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

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{(x + 1) (2 x - 3)} - \frac{3}{2} < 0$

Step 2.1.3

Cancel the common factor of $2 x - 3$ .

Tap for more steps...

Step 2.1.3.1

Cancel the common factor.

$\frac{(2 x - 3) (4 x^{2} + 6 x + 9)}{(x + 1) (2 x - 3)} - \frac{3}{2} < 0$

Step 2.1.3.2

Rewrite the expression.

$\frac{4 x^{2} + 6 x + 9}{x + 1} - \frac{3}{2} < 0$

Step 2.2

To write $\frac{4 x^{2} + 6 x + 9}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 x^{2} + 6 x + 9}{x + 1} \cdot \frac{2}{2} - \frac{3}{2} < 0$

Step 2.3

To write $- \frac{3}{2}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{4 x^{2} + 6 x + 9}{x + 1} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{x + 1}{x + 1} < 0$

Step 2.4

Write each expression with a common denominator of $(x + 1) \cdot 2$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 2.4.1

Multiply $\frac{4 x^{2} + 6 x + 9}{x + 1}$ by $\frac{2}{2}$ .

$\frac{(4 x^{2} + 6 x + 9) \cdot 2}{(x + 1) \cdot 2} - \frac{3}{2} \cdot \frac{x + 1}{x + 1} < 0$

Step 2.4.2

Multiply $\frac{3}{2}$ by $\frac{x + 1}{x + 1}$ .

$\frac{(4 x^{2} + 6 x + 9) \cdot 2}{(x + 1) \cdot 2} - \frac{3 (x + 1)}{2 (x + 1)} < 0$

Step 2.4.3

Reorder the factors of $(x + 1) \cdot 2$ .

$\frac{(4 x^{2} + 6 x + 9) \cdot 2}{2 (x + 1)} - \frac{3 (x + 1)}{2 (x + 1)} < 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{(4 x^{2} + 6 x + 9) \cdot 2 - 3 (x + 1)}{2 (x + 1)} < 0$

Step 2.6

Simplify the numerator.

Tap for more steps...

Step 2.6.1

Apply the distributive property.

$\frac{4 x^{2} \cdot 2 + 6 x \cdot 2 + 9 \cdot 2 - 3 (x + 1)}{2 (x + 1)} < 0$

Step 2.6.2

Simplify.

Tap for more steps...

Step 2.6.2.1

Multiply $2$ by $4$ .

$\frac{8 x^{2} + 6 x \cdot 2 + 9 \cdot 2 - 3 (x + 1)}{2 (x + 1)} < 0$

Step 2.6.2.2

Multiply $2$ by $6$ .

$\frac{8 x^{2} + 12 x + 9 \cdot 2 - 3 (x + 1)}{2 (x + 1)} < 0$

Step 2.6.2.3

Multiply $9$ by $2$ .

$\frac{8 x^{2} + 12 x + 18 - 3 (x + 1)}{2 (x + 1)} < 0$

Step 2.6.3

Apply the distributive property.

$\frac{8 x^{2} + 12 x + 18 - 3 x - 3 \cdot 1}{2 (x + 1)} < 0$

Step 2.6.4

Multiply $- 3$ by $1$ .

$\frac{8 x^{2} + 12 x + 18 - 3 x - 3}{2 (x + 1)} < 0$

Step 2.6.5

Subtract $3 x$ from $12 x$ .

$\frac{8 x^{2} + 9 x + 18 - 3}{2 (x + 1)} < 0$

Step 2.6.6

Subtract $3$ from $18$ .

$\frac{8 x^{2} + 9 x + 15}{2 (x + 1)} < 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$8 x^{2} + 9 x + 15 = 0$

$x + 1 = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 8$ , $b = 9$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (8 \cdot 15)}}{2 \cdot 8}$

Step 6

Simplify.

Tap for more steps...

Step 6.1

Simplify the numerator.

Tap for more steps...

Step 6.1.1

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 8 \cdot 15}}{2 \cdot 8}$

Step 6.1.2

Multiply $- 4 \cdot 8 \cdot 15$ .

Tap for more steps...

Step 6.1.2.1

Multiply $- 4$ by $8$ .

$x = \frac{- 9 \pm \sqrt{81 - 32 \cdot 15}}{2 \cdot 8}$

Step 6.1.2.2

Multiply $- 32$ by $15$ .

$x = \frac{- 9 \pm \sqrt{81 - 480}}{2 \cdot 8}$

Step 6.1.3

Subtract $480$ from $81$ .

$x = \frac{- 9 \pm \sqrt{- 399}}{2 \cdot 8}$

Step 6.1.4

Rewrite $- 399$ as $- 1 (399)$ .

$x = \frac{- 9 \pm \sqrt{- 1 \cdot 399}}{2 \cdot 8}$

Step 6.1.5

Rewrite $\sqrt{- 1 (399)}$ as $\sqrt{- 1} \cdot \sqrt{399}$ .

$x = \frac{- 9 \pm \sqrt{- 1} \cdot \sqrt{399}}{2 \cdot 8}$

Step 6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 9 \pm i \sqrt{399}}{2 \cdot 8}$

Step 6.2

Multiply $2$ by $8$ .

$x = \frac{- 9 \pm i \sqrt{399}}{16}$

Step 7

The final answer is the combination of both solutions.

$x = - \frac{9 - i \sqrt{399}}{16}, - \frac{9 + i \sqrt{399}}{16}$

Step 8

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = - \frac{9 - i \sqrt{399}}{16}, - \frac{9 + i \sqrt{399}}{16}$

$x = - 1$

Step 10

Find the domain of $\frac{8 x^{2} + 9 x + 15}{2 (x + 1)}$ .

Tap for more steps...

Step 10.1

Set the denominator in $\frac{8 x^{2} + 9 x + 15}{2 (x + 1)}$ equal to $0$ to find where the expression is undefined.

$2 (x + 1) = 0$

Step 10.2

Solve for $x$ .

Tap for more steps...

Step 10.2.1

Divide each term in $2 (x + 1) = 0$ by $2$ and simplify.

Tap for more steps...

Step 10.2.1.1

Divide each term in $2 (x + 1) = 0$ by $2$ .

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

Step 10.2.1.2

Simplify the left side.

Tap for more steps...

Step 10.2.1.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 10.2.1.2.1.1

Cancel the common factor.

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

Step 10.2.1.2.1.2

Divide $x + 1$ by $1$ .

$x + 1 = \frac{0}{2}$

Step 10.2.1.3

Simplify the right side.

Tap for more steps...

Step 10.2.1.3.1

Divide $0$ by $2$ .

$x + 1 = 0$

Step 10.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 10.3

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

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

Step 11

The solution consists of all of the true intervals.

$x < - 1$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 1$

Interval Notation:

$(- \infty, - 1)$

Step 13