Precalculus Examples

$x^{3} + 6 x^{2} + 9 x < 0$

Step 1

Convert the inequality to an equation.

$x^{3} + 6 x^{2} + 9 x = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $x$ out of $x^{3} + 6 x^{2} + 9 x$ .

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Step 2.1.1

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

$x \cdot x^{2} + 6 x^{2} + 9 x = 0$

Step 2.1.2

Factor $x$ out of $6 x^{2}$ .

$x \cdot x^{2} + x (6 x) + 9 x = 0$

Step 2.1.3

Factor $x$ out of $9 x$ .

$x \cdot x^{2} + x (6 x) + x \cdot 9 = 0$

Step 2.1.4

Factor $x$ out of $x \cdot x^{2} + x (6 x)$ .

$x (x^{2} + 6 x) + x \cdot 9 = 0$

Step 2.1.5

Factor $x$ out of $x (x^{2} + 6 x) + x \cdot 9$ .

$x (x^{2} + 6 x + 9) = 0$

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

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

$x (x^{2} + 6 x + 3^{2}) = 0$

Step 2.2.2

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

$6 x = 2 \cdot x \cdot 3$

Step 2.2.3

Rewrite the polynomial.

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

Step 2.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Step 3

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

$x = 0$

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

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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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

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

$x = 0, - 3$

Step 7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$x > 0$

Step 8

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

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Step 8.1

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

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Step 8.1.1

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

$x = - 6$

Step 8.1.2

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

${(- 6)}^{3} + 6 {(- 6)}^{2} + 9 (- 6) < 0$

Step 8.1.3

The left side $- 54$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 8.2

Test a value on the interval $- 3 < x < 0$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval $- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 8.2.2

Replace $x$ with $- 2$ in the original inequality.

${(- 2)}^{3} + 6 {(- 2)}^{2} + 9 (- 2) < 0$

Step 8.2.3

The left side $- 2$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 8.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 8.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 8.3.2

Replace $x$ with $2$ in the original inequality.

${(2)}^{3} + 6 {(2)}^{2} + 9 (2) < 0$

Step 8.3.3

The left side $50$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 8.4

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

$x < - 3$ True

$- 3 < x < 0$ True

$x > 0$ False

$x < - 3$ True

$- 3 < x < 0$ True

$x > 0$ False

Step 9

The solution consists of all of the true intervals.

$x < - 3$ or $- 3 < x < 0$

Step 10

Convert the inequality to interval notation.

$(- \infty, - 3) \cup (- 3, 0)$

Step 11