Find the Roots (Zeros) (x^3+9)(x^2-4)=0

$(x^{3} + 9) (x^{2} - 4) = 0$

Step 1

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

$x^{3} + 9 = 0$

$x^{2} - 4 = 0$

Step 2

Set $x^{3} + 9$ equal to $0$ and solve for $x$ .

Set $x^{3} + 9$ equal to $0$ .

$x^{3} + 9 = 0$

Solve $x^{3} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{3} = - 9$

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

$x = \sqrt[3]{- 9}$

Simplify $\sqrt[3]{- 9}$ .

Rewrite $- 9$ as ${(- 1)}^{3} \cdot 9$ .

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

$x = \sqrt[3]{- 1 (9)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x = \sqrt[3]{{(- 1)}^{3} \cdot 9}$

Pull terms out from under the radical.

$x = - 1 \sqrt[3]{9}$

Rewrite $- 1 \sqrt[3]{9}$ as $- \sqrt[3]{9}$ .

$x = - \sqrt[3]{9}$

Step 3

Set $x^{2} - 4$ equal to $0$ and solve for $x$ .

Set $x^{2} - 4$ equal to $0$ .

$x^{2} - 4 = 0$

Solve $x^{2} - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$x^{2} = 4$

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

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

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

$x = \pm 2$

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.

$x = 2$

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

$x = - 2$

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

$x = 2, - 2$

Step 4

The final solution is all the values that make $(x^{3} + 9) (x^{2} - 4) = 0$ true.

$x = - \sqrt[3]{9}, 2, - 2$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \sqrt[3]{9}, 2, - 2$

Decimal Form:

$x = - 2.08008382 \dots, 2, - 2$

Step 6

Find the Roots (Zeros) (x^3+9)(x^2-4)=0

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