Solve the Inequality for x 3x^2+14x+5<=3x-3

$3 x^{2} + 14 x + 5 \leq 3 x - 3$

Step 1

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

Subtract $3 x$ from both sides of the inequality.

$3 x^{2} + 14 x + 5 - 3 x \leq - 3$

Subtract $3 x$ from $14 x$ .

$3 x^{2} + 11 x + 5 \leq - 3$

Step 2

Convert the inequality to an equation.

$3 x^{2} + 11 x + 5 = - 3$

Step 3

Add $3$ to both sides of the equation.

$3 x^{2} + 11 x + 5 + 3 = 0$

Step 4

Add $5$ and $3$ .

$3 x^{2} + 11 x + 8 = 0$

Step 5

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 8 = 24$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$3 x^{2} + 11 (x) + 8 = 0$

Rewrite $11$ as $3$ plus $8$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$3 x (x + 1) + 8 (x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x + 1) (3 x + 8) = 0$

Step 6

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

$x + 1 = 0$

$3 x + 8 = 0$

Step 7

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

Set $3 x + 8$ equal to $0$ and solve for $x$ .

Set $3 x + 8$ equal to $0$ .

$3 x + 8 = 0$

Solve $3 x + 8 = 0$ for $x$ .

Subtract $8$ from both sides of the equation.

$3 x = - 8$

Divide each term in $3 x = - 8$ by $3$ and simplify.

Divide each term in $3 x = - 8$ by $3$ .

$\frac{3 x}{3} = \frac{- 8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 8}{3}$

Divide $x$ by $1$ .

$x = \frac{- 8}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{8}{3}$

Step 9

The final solution is all the values that make $(x + 1) (3 x + 8) = 0$ true.

$x = - 1, - \frac{8}{3}$

Step 10

Use each root to create test intervals.

$x < - \frac{8}{3}$

$- \frac{8}{3} < x < - 1$

$x > - 1$

Step 11

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 $x < - \frac{8}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{8}{3}$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$3 {(- 5)}^{2} + 14 (- 5) + 5 \leq 3 (- 5) - 3$

The left side $10$ is greater than the right side $- 18$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{8}{3} < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{8}{3} < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

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

$3 {(- 2)}^{2} + 14 (- 2) + 5 \leq 3 (- 2) - 3$

The left side $- 11$ is less than the right side $- 9$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$3 {(0)}^{2} + 14 (0) + 5 \leq 3 (0) - 3$

The left side $5$ is greater than the right side $- 3$ , which means that the given statement is false.

False

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

$x < - \frac{8}{3}$ False

$- \frac{8}{3} < x < - 1$ True

$x > - 1$ False

$x < - \frac{8}{3}$ False

$- \frac{8}{3} < x < - 1$ True

$x > - 1$ False

Step 12

The solution consists of all of the true intervals.

$- \frac{8}{3} \leq x \leq - 1$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- \frac{8}{3} \leq x \leq - 1$

Interval Notation:

$[- \frac{8}{3}, - 1]$

Step 14

Solve the Inequality for x 3x^2+14x+5<=3x-3

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