Convert to Interval Notation x^3-3x^2-9x+27<0

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

Step 1

Convert the inequality to an equation.

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

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.

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

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 $1$ .

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

Step 3

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

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

$x + 3 = 0$

Step 4

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 $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

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 6

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

$x = 3, - 3$

Step 7

The solution consists of all of the true intervals.

$x < - 3$

Step 8

Convert the inequality to interval notation.

$(- \infty, - 3)$

Step 9

Convert to Interval Notation x^3-3x^2-9x+27<0

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