Find the Roots (Zeros) x^5+2x^3+x=0

$x^{5} + 2 x^{3} + x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{5} + 2 x^{3} + x$ .

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

$x \cdot x^{4} + 2 x^{3} + x = 0$

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

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

$x \cdot x^{4} + x (2 x^{2}) + x \cdot 1 = 0$

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

$x (x^{4} + 2 x^{2}) + x \cdot 1 = 0$

Factor $x$ out of $x (x^{4} + 2 x^{2}) + x \cdot 1$ .

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

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

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

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

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

Factor using the perfect square rule.

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

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

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

$2 u = 2 \cdot u \cdot 1$

Rewrite the polynomial.

$x (u^{2} + 2 \cdot u \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

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

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

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

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

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

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

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

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

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

$x^{2} + 1 = 0$

Solve 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.

$x = i, - i$

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

$x = 0, i, - i$

Find the Roots (Zeros) x^5+2x^3+x=0

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