Find the Roots (Zeros) x^4-x^5

$x^{4} - x^{5}$

Step 1

Set $x^{4} - x^{5}$ equal to $0$ .

$x^{4} - x^{5} = 0$

Step 2

Solve for $x$ .

Factor $x^{4}$ out of $x^{4} - x^{5}$ .

Multiply by $1$ .

$x^{4} \cdot 1 - x^{5} = 0$

Factor $x^{4}$ out of $- x^{5}$ .

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

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

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

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

$x^{4} = 0$

$1 - x = 0$

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

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

$x^{4} = 0$

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

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

$x = \pm \sqrt[4]{0}$

Simplify $\pm \sqrt[4]{0}$ .

Rewrite $0$ as $0^{4}$ .

$x = \pm \sqrt[4]{0^{4}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

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

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The final solution is all the values that make $x^{4} (1 - x) = 0$ true.

$x = 0, 1$

Step 3

Find the Roots (Zeros) x^4-x^5

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