Algebra Examples

$x^{3} - 8 x^{2} - x + 8 \geq 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 8 x^{2} - x + 8 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor out the greatest common factor from each group.

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Step 2.1.1

Group the first two terms and the last two terms.

$(x^{3} - 8 x^{2}) - x + 8 = 0$

Step 2.1.2

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

$x^{2} (x - 8) - (x - 8) = 0$

Step 2.2

Factor the polynomial by factoring out the greatest common factor, $x - 8$ .

$(x - 8) (x^{2} - 1) = 0$

Step 2.3

Rewrite $1$ as $1^{2}$ .

$(x - 8) (x^{2} - 1^{2}) = 0$

Step 2.4

Factor.

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Step 2.4.1

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 $b = 1$ .

$(x - 8) ((x + 1) (x - 1)) = 0$

Step 2.4.2

Remove unnecessary parentheses.

$(x - 8) (x + 1) (x - 1) = 0$

Step 3

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

$x - 8 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 4

Set $x - 8$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 4.2

Add $8$ to both sides of the equation.

$x = 8$

Step 5

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

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Step 5.1

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

$x + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

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

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Step 6.1

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

$x - 1 = 0$

Step 6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 7

The final solution is all the values that make $(x - 8) (x + 1) (x - 1) = 0$ true.

$x = 8, - 1, 1$

Step 8

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 8$

$x > 8$

Step 9

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 9.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 9.1.2

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{3} - 8 {(- 4)}^{2} - (- 4) + 8 \geq 0$

Step 9.1.3

The left side $- 180$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

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

${(0)}^{3} - 8 {(0)}^{2} - (0) + 8 \geq 0$

Step 9.2.3

The left side $8$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $1 < x < 8$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $1 < x < 8$ and see if this value makes the original inequality true.

$x = 4$

Step 9.3.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{3} - 8 {(4)}^{2} - (4) + 8 \geq 0$

Step 9.3.3

The left side $- 60$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.4

Test a value on the interval $x > 8$ to see if it makes the inequality true.

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Step 9.4.1

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Step 9.4.2

Replace $x$ with $10$ in the original inequality.

${(10)}^{3} - 8 {(10)}^{2} - (10) + 8 \geq 0$

Step 9.4.3

The left side $198$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.5

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

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 8$ False

$x > 8$ True

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 8$ False

$x > 8$ True

Step 10

The solution consists of all of the true intervals.

$- 1 \leq x \leq 1$ or $x \geq 8$

Step 11

Convert the inequality to interval notation.

$[- 1, 1] \cup [8, \infty)$

Step 12