Solve for x 3(x^2-1)>-8x

$3 (x^{2} - 1) > - 8 x$

Step 1

Simplify $3 (x^{2} - 1)$ .

Rewrite.

$0 + 0 + 3 (x^{2} - 1) > - 8 x$

Simplify by adding zeros.

$3 (x^{2} - 1) > - 8 x$

Apply the distributive property.

$3 x^{2} + 3 \cdot - 1 > - 8 x$

Multiply $3$ by $- 1$ .

$3 x^{2} - 3 > - 8 x$

Step 2

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

$3 x^{2} - 3 + 8 x > 0$

Step 3

Convert the inequality to an equation.

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

Step 4

Factor by grouping.

Reorder terms.

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

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 $a \cdot c = 3 \cdot - 3 = - 9$ and whose sum is $b = 8$ .

Factor $8$ out of $8 x$ .

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

Rewrite $8$ as $- 1$ plus $9$

$3 x^{2} + (- 1 + 9) x - 3 = 0$

Apply the distributive property.

$3 x^{2} - 1 x + 9 x - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) + 9 x - 3 = 0$

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

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

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

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

Step 5

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

$3 x - 1 = 0$

$x + 3 = 0$

Step 6

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 $x$ .

Add $1$ to 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 $1$ .

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

Step 7

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 8

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

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

Step 9

Use each root to create test intervals.

$x < - 3$

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

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

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$3 ({(- 6)}^{2} - 1) > - 8 \cdot - 6$

The left side $105$ is greater than the right side $48$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$3 ({(0)}^{2} - 1) > - 8 \cdot 0$

The left side $- 3$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

$x = 3$

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

$3 ({(3)}^{2} - 1) > - 8 \cdot 3$

The left side $24$ is greater than the right side $- 24$ , which means that the given statement is always true.

True

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

$x < - 3$ True

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

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

$x < - 3$ True

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

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

Step 11

The solution consists of all of the true intervals.

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

Step 12

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

$(- \infty, - 3) \cup (\frac{1}{3}, \infty)$

Step 13

Solve for x 3(x^2-1)>-8x

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