Find the Roots (Zeros) x^5-6x^3+9x

$x^{5} - 6 x^{3} + 9 x$

Set $x^{5} - 6 x^{3} + 9 x$ equal to $0$ .

$x^{5} - 6 x^{3} + 9 x = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{5} - 6 x^{3} + 9 x$ .

Factor $x$ out of $x^{5}$ .

$x \cdot x^{4} - 6 x^{3} + 9 x = 0$

Factor $x$ out of $- 6 x^{3}$ .

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

Factor $x$ out of $9 x$ .

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

Factor $x$ out of $x \cdot x^{4} + x (- 6 x^{2})$ .

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

Factor $x$ out of $x (x^{4} - 6 x^{2}) + x \cdot 9$ .

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

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

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

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

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

Factor using the perfect square rule.

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

$x (u^{2} - 6 u + 3^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Rewrite the polynomial.

$x (u^{2} - 2 \cdot u \cdot 3 + 3^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

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

Replace all occurrences of $u$ with $x^{2}$ .

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

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set ${(x^{2} - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x^{2} - 3)}^{2}$ equal to $0$ .

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

Solve ${(x^{2} - 3)}^{2} = 0$ for $x$ .

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

$x^{2} - 3 = 0$

Solve for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

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

$x = 0, \sqrt{3}, - \sqrt{3}$

The result can be shown in multiple forms.

Exact Form:

$x = 0, \sqrt{3}, - \sqrt{3}$

Decimal Form:

$x = 0, 1.73205080 \dots, - 1.73205080 \dots$

Find the Roots (Zeros) x^5-6x^3+9x

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