Identify the Zeros and Their Multiplicities (x^2-9)^4

${(x^{2} - 9)}^{4}$

Step 1

Set ${(x^{2} - 9)}^{4}$ equal to $0$ .

${(x^{2} - 9)}^{4} = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Rewrite $9$ as $3^{2}$ .

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

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 $b = 3$ .

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

Apply the product rule to $(x + 3) (x - 3)$ .

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

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

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

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

Set ${(x + 3)}^{4}$ equal to $0$ and solve for $x$ .

Set ${(x + 3)}^{4}$ equal to $0$ .

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

Solve ${(x + 3)}^{4} = 0$ for $x$ .

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set ${(x - 3)}^{4}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{4}$ equal to $0$ .

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

Solve ${(x - 3)}^{4} = 0$ for $x$ .

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = - 3$ (Multiplicity of $4$ )

$x = 3$ (Multiplicity of $4$ )

$x = - 3$ (Multiplicity of $4$ )

$x = 3$ (Multiplicity of $4$ )

Step 3

Identify the Zeros and Their Multiplicities (x^2-9)^4

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