Solve the Inequality for x -3x^2-6x+31<7

$- 3 x^{2} - 6 x + 31 < 7$

Step 1

Convert the inequality to an equation.

$- 3 x^{2} - 6 x + 31 = 7$

Step 2

Subtract $7$ from both sides of the equation.

$- 3 x^{2} - 6 x + 31 - 7 = 0$

Step 3

Subtract $7$ from $31$ .

$- 3 x^{2} - 6 x + 24 = 0$

Step 4

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} - 6 x + 24$ .

Factor $- 3$ out of $- 3 x^{2}$ .

$- 3 x^{2} - 6 x + 24 = 0$

Factor $- 3$ out of $- 6 x$ .

$- 3 x^{2} - 3 (2 x) + 24 = 0$

Factor $- 3$ out of $24$ .

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

Factor $- 3$ out of $- 3 (x^{2}) - 3 (2 x)$ .

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

Factor $- 3$ out of $- 3 (x^{2} + 2 x) - 3 (- 8)$ .

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

Factor.

Factor $x^{2} + 2 x - 8$ 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 $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$- 3 ((x - 2) (x + 4)) = 0$

Remove unnecessary parentheses.

$- 3 (x - 2) (x + 4) = 0$

Step 5

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

$x - 2 = 0$

$x + 4 = 0$

Step 6

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 7

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

The final solution is all the values that make $- 3 (x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

Step 9

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 2$

$x > 2$

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 $x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

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

$- 3 {(- 6)}^{2} - 6 \cdot - 6 + 31 < 7$

The left side $- 41$ is less than the right side $7$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$- 3 {(0)}^{2} - 6 \cdot 0 + 31 < 7$

The left side $31$ is not less than the right side $7$ , which means that the given statement is false.

False

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

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

$x = 4$

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

$- 3 {(4)}^{2} - 6 \cdot 4 + 31 < 7$

The left side $- 41$ is less than the right side $7$ , which means that the given statement is always true.

True

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

$x < - 4$ True

$- 4 < x < 2$ False

$x > 2$ True

$x < - 4$ True

$- 4 < x < 2$ False

$x > 2$ True

Step 11

The solution consists of all of the true intervals.

$x < - 4$ or $x > 2$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 4 or x > 2$

Interval Notation:

$(- \infty, - 4) \cup (2, \infty)$

Step 13

Solve the Inequality for x -3x^2-6x+31<7

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