Convert to Interval Notation x^3+x^2+9x+9<0

$x^{3} + x^{2} + 9 x + 9 < 0$

Step 1

Convert the inequality to an equation.

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

Step 2

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Step 3

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

$x + 1 = 0$

$x^{2} + 9 = 0$

Step 4

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 5

Set $x^{2} + 9$ equal to $0$ and solve for $x$ .

Set $x^{2} + 9$ equal to $0$ .

$x^{2} + 9 = 0$

Solve $x^{2} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 9}$

Simplify $\pm \sqrt{- 9}$ .

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i, - 3 i$

Step 6

The final solution is all the values that make $(x + 1) (x^{2} + 9) = 0$ true.

$x = - 1, 3 i, - 3 i$

Step 7

The solution consists of all of the true intervals.

$x < - 1$

Step 8

Convert the inequality to interval notation.

$(- \infty, - 1)$

Step 9

Convert to Interval Notation x^3+x^2+9x+9<0

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