Basic Math Examples

$y^{5} + 16 y = 0$

Step 1

Factor $y$ out of $y^{5} + 16 y$ .

Tap for more steps...

Step 1.1

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

$y \cdot y^{4} + 16 y = 0$

Step 1.2

Factor $y$ out of $16 y$ .

$y \cdot y^{4} + y \cdot 16 = 0$

Step 1.3

Factor $y$ out of $y \cdot y^{4} + y \cdot 16$ .

$y (y^{4} + 16) = 0$

Step 2

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

$y = 0$

$y^{4} + 16 = 0$

Step 3

Set $y$ equal to $0$ .

$y = 0$

Step 4

Set $y^{4} + 16$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 4.1

Set $y^{4} + 16$ equal to $0$ .

$y^{4} + 16 = 0$

Step 4.2

Solve $y^{4} + 16 = 0$ for $y$ .

Tap for more steps...

Step 4.2.1

Subtract $16$ from both sides of the equation.

$y^{4} = - 16$

Step 4.2.2

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

$y = \pm \sqrt[4]{- 16}$

Step 4.2.3

Simplify $\pm \sqrt[4]{- 16}$ .

Tap for more steps...

Step 4.2.3.1

Rewrite $- 16$ as $2^{4} \cdot - 1$ .

Tap for more steps...

Step 4.2.3.1.1

Factor $16$ out of $- 16$ .

$y = \pm \sqrt[4]{16 (- 1)}$

Step 4.2.3.1.2

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

$y = \pm \sqrt[4]{2^{4} \cdot - 1}$

Step 4.2.3.2

Pull terms out from under the radical.

$y = \pm 2 \sqrt[4]{- 1}$

Step 4.2.4

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

Tap for more steps...

Step 4.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$y = 2 \sqrt[4]{- 1}$

Step 4.2.4.2

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

$y = - 2 \sqrt[4]{- 1}$

Step 4.2.4.3

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

$y = 2 \sqrt[4]{- 1}, - 2 \sqrt[4]{- 1}$

Step 5

The final solution is all the values that make $y (y^{4} + 16) = 0$ true.

$y = 0, 2 \sqrt[4]{- 1}, - 2 \sqrt[4]{- 1}$