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

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

Step 1

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

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

Step 2

Solve for $x$ .

Factor.

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

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

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 $b = 2$ .

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

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

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

$x + 2 = 0$

$x - 2 = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

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

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 2$ equal to $0$ and solve for $x$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ 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 $- 3 {(x - 1)}^{2} (x^{2} - 4) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 1$ (Multiplicity of $2$ )

$x = - 2$ (Multiplicity of $1$ )

$x = 2$ (Multiplicity of $1$ )

$x = 1$ (Multiplicity of $2$ )

$x = - 2$ (Multiplicity of $1$ )

$x = 2$ (Multiplicity of $1$ )

Step 3

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

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