Solve the Inequality for x 3x^2+16x<-5

$3 x^{2} + 16 x < - 5$

Convert the inequality to an equation.

$3 x^{2} + 16 x = - 5$

Add $5$ to both sides of the equation.

$3 x^{2} + 16 x + 5 = 0$

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 $a \cdot c = 3 \cdot 5 = 15$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$3 x^{2} + 16 (x) + 5 = 0$

Rewrite $16$ as $1$ plus $15$

$3 x^{2} + (1 + 15) x + 5 = 0$

Apply the distributive property.

$3 x^{2} + 1 x + 15 x + 5 = 0$

Multiply $x$ by $1$ .

$3 x^{2} + x + 15 x + 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + x) + 15 x + 5 = 0$

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

$x (3 x + 1) + 5 (3 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$(3 x + 1) (x + 5) = 0$

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

$3 x + 1 = 0$

$x + 5 = 0$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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.

$x = - 5$

The final solution is all the values that make $(3 x + 1) (x + 5) = 0$ true.

$x = - \frac{1}{3}, - 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - \frac{1}{3}$

$x > - \frac{1}{3}$

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.

$3 {(- 8)}^{2} + 16 (- 8) < - 5$

The left side $64$ is not less than the right side $- 5$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < - \frac{1}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - \frac{1}{3}$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

$3 {(- 3)}^{2} + 16 (- 3) < - 5$

The left side $- 21$ is less than the right side $- 5$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$3 {(0)}^{2} + 16 (0) < - 5$

The left side $0$ is not less than the right side $- 5$ , which means that the given statement is false.

False

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

$x < - 5$ False

$- 5 < x < - \frac{1}{3}$ True

$x > - \frac{1}{3}$ False

$x < - 5$ False

$- 5 < x < - \frac{1}{3}$ True

$x > - \frac{1}{3}$ False

The solution consists of all of the true intervals.

$- 5 < x < - \frac{1}{3}$

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x < - \frac{1}{3}$

Interval Notation:

$(- 5, - \frac{1}{3})$

Solve the Inequality for x 3x^2+16x<-5

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