Find the Roots (Zeros) (x^2+1)(x^3+2x)(x^2-64)=0

$(x^{2} + 1) (x^{3} + 2 x) (x^{2} - 64) = 0$

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

$x^{2} + 1 = 0$

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

$x^{2} - 64 = 0$

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

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

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

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

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

Factor $x$ out of $x^{3} + 2 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} + 2 x = 0$

Factor $x$ out of $2 x$ .

$x \cdot x^{2} + x \cdot 2 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot 2$ .

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

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

$x = 0$

$x^{2} + 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x^{2} + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$x^{2} = - 2$

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

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

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{2}$

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

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

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.

$x = i \sqrt{2}$

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

$x = - i \sqrt{2}$

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

$x = i \sqrt{2}, - i \sqrt{2}$

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

$x = 0, i \sqrt{2}, - i \sqrt{2}$

Set $x^{2} - 64$ equal to $0$ and solve for $x$ .

Set $x^{2} - 64$ equal to $0$ .

$x^{2} - 64 = 0$

Solve $x^{2} - 64 = 0$ for $x$ .

Add $64$ to both sides of the equation.

$x^{2} = 64$

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

$x = \pm \sqrt{64}$

Simplify $\pm \sqrt{64}$ .

Rewrite $64$ as $8^{2}$ .

$x = \pm \sqrt{8^{2}}$

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

$x = \pm 8$

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.

$x = 8$

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

$x = - 8$

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

$x = 8, - 8$

The final solution is all the values that make $(x^{2} + 1) (x^{3} + 2 x) (x^{2} - 64) = 0$ true.

$x = i, - i, 0, i \sqrt{2}, - i \sqrt{2}, 8, - 8$

Find the Roots (Zeros) (x^2+1)(x^3+2x)(x^2-64)=0

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