Solve the Inequality for x x^3+3x^2-9x-27>0

$x^{3} + 3 x^{2} - 9 x - 27 > 0$

Convert the inequality to an equation.

$x^{3} + 3 x^{2} - 9 x - 27 = 0$

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.

$(x^{3} + 3 x^{2}) - 9 x - 27 = 0$

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

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

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

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

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

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

Remove unnecessary parentheses.

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

Combine exponents.

Raise $x + 3$ to the power of $1$ .

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

Raise $x + 3$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

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

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

$x - 3 = 0$

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 3)}^{2}$ equal to $0$ .

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

Solve ${(x + 3)}^{2} = 0$ for $x$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make ${(x + 3)}^{2} (x - 3) = 0$ true.

$x = - 3, 3$

The solution consists of all of the true intervals.

$x > 3$

The result can be shown in multiple forms.

Inequality Form:

$x > 3$

Interval Notation:

$(3, \infty)$

Solve the Inequality for x x^3+3x^2-9x-27>0

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