Solve for x x^6-9x^4-x^2+9=0

$x^{6} - 9 x^{4} - x^{2} + 9 = 0$

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.

$(x^{6} - 9 x^{4}) - x^{2} + 9 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $x^{2} - 9$ .

$(x^{2} - 9) (x^{4} - 1) = 0$

Rewrite $9$ as $3^{2}$ .

$(x^{2} - 3^{2}) (x^{4} - 1) = 0$

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 $b = 3$ .

$(x + 3) (x - 3) (x^{4} - 1) = 0$

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

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

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

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

Factor.

Simplify.

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

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

$x + 3 = 0$

$x - 3 = 0$

$x^{2} + 1 = 0$

$x + 1 = 0$

$x - 1 = 0$

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.

$x = - 3$

Set $x - 3$ equal to $0$ and solve for $x$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = - 3, 3, i, - i, - 1, 1$

Solve for x x^6-9x^4-x^2+9=0

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