Solve the Inequality for x 3x^2+40x<-8x-189

$3 x^{2} + 40 x < - 8 x - 189$

Move all terms containing $x$ to the left side of the inequality.

Add $8 x$ to both sides of the inequality.

$3 x^{2} + 40 x + 8 x < - 189$

Add $40 x$ and $8 x$ .

$3 x^{2} + 48 x < - 189$

Convert the inequality to an equation.

$3 x^{2} + 48 x = - 189$

Add $189$ to both sides of the equation.

$3 x^{2} + 48 x + 189 = 0$

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} + 48 x + 189$ .

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

$3 (x^{2}) + 48 x + 189 = 0$

Factor $3$ out of $48 x$ .

$3 (x^{2}) + 3 (16 x) + 189 = 0$

Factor $3$ out of $189$ .

$3 x^{2} + 3 (16 x) + 3 \cdot 63 = 0$

Factor $3$ out of $3 x^{2} + 3 (16 x)$ .

$3 (x^{2} + 16 x) + 3 \cdot 63 = 0$

Factor $3$ out of $3 (x^{2} + 16 x) + 3 \cdot 63$ .

$3 (x^{2} + 16 x + 63) = 0$

Factor.

Factor $x^{2} + 16 x + 63$ 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 $63$ and whose sum is $16$ .

$7, 9$

Write the factored form using these integers.

$3 ((x + 7) (x + 9)) = 0$

Remove unnecessary parentheses.

$3 (x + 7) (x + 9) = 0$

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

$x + 7 = 0$

$x + 9 = 0$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

The final solution is all the values that make $3 (x + 7) (x + 9) = 0$ true.

$x = - 7, - 9$

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 7$

$x > - 7$

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 $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$3 {(- 12)}^{2} + 40 (- 12) < - 8 \cdot - 12 - 189$

The left side $- 48$ is not less than the right side $- 93$ , which means that the given statement is false.

False

Test a value on the interval $- 9 < x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < - 7$ and see if this value makes the original inequality true.

$x = - 8$

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

$3 {(- 8)}^{2} + 40 (- 8) < - 8 \cdot - 8 - 189$

The left side $- 128$ is less than the right side $- 125$ , which means that the given statement is always true.

True

Test a value on the interval $x > - 7$ to see if it makes the inequality true.

Choose a value on the interval $x > - 7$ and see if this value makes the original inequality true.

$x = 0$

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

$3 {(0)}^{2} + 40 (0) < - 8 \cdot 0 - 189$

The left side $0$ is not less than the right side $- 189$ , which means that the given statement is false.

False

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

$x < - 9$ False

$- 9 < x < - 7$ True

$x > - 7$ False

$x < - 9$ False

$- 9 < x < - 7$ True

$x > - 7$ False

The solution consists of all of the true intervals.

$- 9 < x < - 7$

The result can be shown in multiple forms.

Inequality Form:

$- 9 < x < - 7$

Interval Notation:

$(- 9, - 7)$

Solve the Inequality for x 3x^2+40x<-8x-189

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