Identify the Zeros and Their Multiplicities x^5-16x^3=0

$x^{5} - 16 x^{3} = 0$

Step 1

Factor the left side of the equation.

Factor $x^{3}$ out of $x^{5} - 16 x^{3}$ .

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

$x^{3} x^{2} - 16 x^{3} = 0$

Factor $x^{3}$ out of $- 16 x^{3}$ .

$x^{3} x^{2} + x^{3} \cdot - 16 = 0$

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} \cdot - 16$ .

$x^{3} (x^{2} - 16) = 0$

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

$x^{3} (x^{2} - 4^{2}) = 0$

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 $b = 4$ .

$x^{3} ((x + 4) (x - 4)) = 0$

Remove unnecessary parentheses.

$x^{3} (x + 4) (x - 4) = 0$

Step 2

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

$x^{3} = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 3

Set $x^{3}$ equal to $0$ and solve for $x$ .

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Solve $x^{3} = 0$ for $x$ .

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

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

$x = 0$

Step 4

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

The final solution is all the values that make $x^{3} (x + 4) (x - 4) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $3$ )

$x = - 4$ (Multiplicity of $1$ )

$x = 4$ (Multiplicity of $1$ )

Step 7

Identify the Zeros and Their Multiplicities x^5-16x^3=0

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