Identify the Zeros and Their Multiplicities (x^2-16)^2

${(x^{2} - 16)}^{2}$

Step 1

Set ${(x^{2} - 16)}^{2}$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

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

Apply the product rule to $(x + 4) (x - 4)$ .

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

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

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

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

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

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

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

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make ${(x + 4)}^{2} {(x - 4)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 4$ (Multiplicity of $2$ )

$x = 4$ (Multiplicity of $2$ )

$x = - 4$ (Multiplicity of $2$ )

$x = 4$ (Multiplicity of $2$ )

Step 3

Identify the Zeros and Their Multiplicities (x^2-16)^2

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