Identify the Zeros and Their Multiplicities -4x^4(x-1)(x+2)^4

$- 4 x^{4} (x - 1) {(x + 2)}^{4}$

Step 1

Set $- 4 x^{4} (x - 1) {(x + 2)}^{4}$ equal to $0$ .

$- 4 x^{4} (x - 1) {(x + 2)}^{4} = 0$

Step 2

Solve for $x$ .

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

$x^{4} = 0$

$x - 1 = 0$

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

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 $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

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

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

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $- 4 x^{4} (x - 1) {(x + 2)}^{4} = 0$ true. The multiplicity of a root is the number of times the root appears.

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

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

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

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

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

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

Step 3

Identify the Zeros and Their Multiplicities -4x^4(x-1)(x+2)^4

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