Find the Roots (Zeros) x^6-2x^4-4x^2+8=0

$x^{6} - 2 x^{4} - 4 x^{2} + 8 = 0$

Step 1

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.

$(x^{6} - 2 x^{4}) - 4 x^{2} + 8 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $x^{2} - 2$ .

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

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

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

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

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

Factor.

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

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

Remove unnecessary parentheses.

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

Combine exponents.

Raise $x^{2} - 2$ to the power of $1$ .

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

Raise $x^{2} - 2$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Step 2

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

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

$x^{2} + 2 = 0$

Step 3

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

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

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

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

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

$x^{2} - 2 = 0$

Solve for $x$ .

Add $2$ to 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.

$x = \pm \sqrt{2}$

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.

$x = \sqrt{2}$

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

$x = - \sqrt{2}$

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

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

Step 4

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.

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

Step 5

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

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

Step 6

Find the Roots (Zeros) x^6-2x^4-4x^2+8=0

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