Identify the Zeros and Their Multiplicities q(x)=-2x^4(x+1)^3(x-2)^2

$q (x) = - 2 x^{4} {(x + 1)}^{3} {(x - 2)}^{2}$

Step 1

Set $- 2 x^{4} {(x + 1)}^{3} {(x - 2)}^{2}$ equal to $0$ .

$- 2 x^{4} {(x + 1)}^{3} {(x - 2)}^{2} = 0$

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 $0$ .

$x^{4} = 0$

${(x + 1)}^{3} = 0$

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

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 ${(x + 1)}^{3}$ equal to $0$ and solve for $x$ .

Set ${(x + 1)}^{3}$ equal to $0$ .

${(x + 1)}^{3} = 0$

Solve ${(x + 1)}^{3} = 0$ for $x$ .

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

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

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $- 2 x^{4} {(x + 1)}^{3} {(x - 2)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

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

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

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

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

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

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

Step 3

Identify the Zeros and Their Multiplicities q(x)=-2x^4(x+1)^3(x-2)^2

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