Identify the Zeros and Their Multiplicities f(x)=-3x^4(x^2-9)

$f (x) = - 3 x^{4} (x^{2} - 9)$

Step 1

Set $- 3 x^{4} (x^{2} - 9)$ equal to $0$ .

$- 3 x^{4} (x^{2} - 9) = 0$

Step 2

Solve for $x$ .

Factor.

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

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

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 $b = 3$ .

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

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

$x^{4} = 0$

$x + 3 = 0$

$x - 3 = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ 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 $- 3 x^{4} (x^{2} - 9) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $4$ )

$x = - 3$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

$x = 0$ (Multiplicity of $4$ )

$x = - 3$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=-3x^4(x^2-9)

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