Solve by Factoring 5x^5-80x=0

$5 x^{5} - 80 x = 0$

Factor $5 x$ out of $5 x^{5} - 80 x$ .

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

$5 x (x^{4}) - 80 x = 0$

Factor $5 x$ out of $- 80 x$ .

$5 x (x^{4}) + 5 x (- 16) = 0$

Factor $5 x$ out of $5 x (x^{4}) + 5 x (- 16)$ .

$5 x (x^{4} - 16) = 0$

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

$5 x ({(x^{2})}^{2} - 16) = 0$

Rewrite $16$ as $4^{2}$ .

$5 x ({(x^{2})}^{2} - 4^{2}) = 0$

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 $b = 4$ .

$5 x ((x^{2} + 4) (x^{2} - 4)) = 0$

Factor.

Simplify.

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

$5 x ((x^{2} + 4) (x^{2} - 2^{2})) = 0$

Factor.

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

$5 x ((x^{2} + 4) ((x + 2) (x - 2))) = 0$

Remove unnecessary parentheses.

$5 x ((x^{2} + 4) (x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$5 x (x^{2} + 4) (x + 2) (x - 2) = 0$

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

$x = 0$

$x^{2} + 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{4}$

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

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

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

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

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

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $5 x (x^{2} + 4) (x + 2) (x - 2) = 0$ true.

$x = 0, 2 i, - 2 i, - 2, 2$

Solve by Factoring 5x^5-80x=0

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