Solve the Inequality for x -x^2(3-x)^3(x+2)>0

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

Step 1

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

$x^{2} = 0$

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

$x + 2 = 0$

Step 2

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 3

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

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

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

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

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

$3 - x = 0$

Solve for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Step 4

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

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

$x = 0, 3, - 2$

Step 6

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$0 < x < 3$

$x > 3$

Step 7

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 $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$- {(- 4)}^{2} {(3 - (- 4))}^{3} ((- 4) + 2) > 0$

The left side $10976$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

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

The left side $- 64$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

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

The left side $- 16$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

$- {(6)}^{2} {(3 - (6))}^{3} ((6) + 2) > 0$

The left side $7776$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 3$ False

$x > 3$ True

$x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 3$ False

$x > 3$ True

Step 8

The solution consists of all of the true intervals.

$x < - 2$ or $x > 3$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or x > 3$

Interval Notation:

$(- \infty, - 2) \cup (3, \infty)$

Step 10

Solve the Inequality for x -x^2(3-x)^3(x+2)>0

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