Convert to Interval Notation x^3+7x^2<-10x

$x^{3} + 7 x^{2} < - 10 x$

Step 1

Add $10 x$ to both sides of the inequality.

$x^{3} + 7 x^{2} + 10 x < 0$

Step 2

Convert the inequality to an equation.

$x^{3} + 7 x^{2} + 10 x = 0$

Step 3

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 7 x^{2} + 10 x$ .

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

$x \cdot x^{2} + 7 x^{2} + 10 x = 0$

Factor $x$ out of $7 x^{2}$ .

$x \cdot x^{2} + x (7 x) + 10 x = 0$

Factor $x$ out of $10 x$ .

$x \cdot x^{2} + x (7 x) + x \cdot 10 = 0$

Factor $x$ out of $x \cdot x^{2} + x (7 x)$ .

$x (x^{2} + 7 x) + x \cdot 10 = 0$

Factor $x$ out of $x (x^{2} + 7 x) + x \cdot 10$ .

$x (x^{2} + 7 x + 10) = 0$

Factor.

Factor $x^{2} + 7 x + 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

$x ((x + 2) (x + 5)) = 0$

Remove unnecessary parentheses.

$x (x + 2) (x + 5) = 0$

Step 4

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

$x = 0$

$x + 2 = 0$

$x + 5 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

The final solution is all the values that make $x (x + 2) (x + 5) = 0$ true.

$x = 0, - 2, - 5$

Step 9

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 2$

$- 2 < x < 0$

$x > 0$

Step 10

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

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

${(- 8)}^{3} + 7 {(- 8)}^{2} < - 10 \cdot - 8$

The left side $- 64$ is less than the right side $80$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

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

${(- 4)}^{3} + 7 {(- 4)}^{2} < - 10 \cdot - 4$

The left side $48$ is not less than the right side $40$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

${(- 1)}^{3} + 7 {(- 1)}^{2} < - 10 \cdot - 1$

The left side $6$ is less than the right side $10$ , which means that the given statement is always true.

True

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

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

$x = 2$

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

${(2)}^{3} + 7 {(2)}^{2} < - 10 \cdot 2$

The left side $36$ is not less than the right side $- 20$ , which means that the given statement is false.

False

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

$x < - 5$ True

$- 5 < x < - 2$ False

$- 2 < x < 0$ True

$x > 0$ False

$x < - 5$ True

$- 5 < x < - 2$ False

$- 2 < x < 0$ True

$x > 0$ False

Step 11

The solution consists of all of the true intervals.

$x < - 5$ or $- 2 < x < 0$

Step 12

Convert the inequality to interval notation.

$(- \infty, - 5) \cup (- 2, 0)$

Step 13

Convert to Interval Notation x^3+7x^2<-10x

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