Precalculus Examples

$x^{3} + 3 x^{2} - 9 x \leq 27$

Step 1

Convert the inequality to an equation.

$x^{3} + 3 x^{2} - 9 x = 27$

Step 2

Subtract $27$ from both sides of the equation.

$x^{3} + 3 x^{2} - 9 x - 27 = 0$

Step 3

Factor the left side of the equation.

Tap for more steps...

Step 3.1

Factor out the greatest common factor from each group.

Tap for more steps...

Step 3.1.1

Group the first two terms and the last two terms.

$(x^{3} + 3 x^{2}) - 9 x - 27 = 0$

Step 3.1.2

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

$x^{2} (x + 3) - 9 (x + 3) = 0$

Step 3.2

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

$(x + 3) (x^{2} - 9) = 0$

Step 3.3

Rewrite $9$ as $3^{2}$ .

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

Step 3.4

Factor.

Tap for more steps...

Step 3.4.1

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 $b = 3$ .

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

Step 3.4.2

Remove unnecessary parentheses.

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

Step 3.5

Combine exponents.

Tap for more steps...

Step 3.5.1

Raise $x + 3$ to the power of $1$ .

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

Step 3.5.2

Raise $x + 3$ to the power of $1$ .

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

Step 3.5.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.5.4

Add $1$ and $1$ .

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

Step 4

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

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

$x - 3 = 0$

Step 5

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 5.1

Set ${(x + 3)}^{2}$ equal to $0$ .

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

Step 5.2

Solve ${(x + 3)}^{2} = 0$ for $x$ .

Tap for more steps...

Step 5.2.1

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

$x + 3 = 0$

Step 5.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

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

Tap for more steps...

Step 6.1

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

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make ${(x + 3)}^{2} (x - 3) = 0$ true.

$x = - 3, 3$

Step 8

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$x > 3$

Step 9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 9.1

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Tap for more steps...

Step 9.1.1

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 9.1.2

Replace $x$ with $- 6$ in the original inequality.

${(- 6)}^{3} + 3 {(- 6)}^{2} - 9 \cdot - 6 \leq 27$

Step 9.1.3

The left side $- 54$ is less than the right side $27$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- 3 < x < 3$ to see if it makes the inequality true.

Tap for more steps...

Step 9.2.1

Choose a value on the interval $- 3 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{3} + 3 {(0)}^{2} - 9 \cdot 0 \leq 27$

Step 9.2.3

The left side $0$ is less than the right side $27$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Tap for more steps...

Step 9.3.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 9.3.2

Replace $x$ with $6$ in the original inequality.

${(6)}^{3} + 3 {(6)}^{2} - 9 \cdot 6 \leq 27$

Step 9.3.3

The left side $270$ is greater than the right side $27$ , which means that the given statement is false.

False

Step 9.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < 3$ True

$x > 3$ False

$x < - 3$ True

$- 3 < x < 3$ True

$x > 3$ False

Step 10

The solution consists of all of the true intervals.

$x \leq - 3$ or $- 3 \leq x \leq 3$

Step 11

Combine the intervals.

$x \leq 3$

Step 12

Convert the inequality to interval notation.

$(- \infty, 3]$

Step 13