Identify the Zeros and Their Multiplicities x^3-4x^2-4x+16

$x^{3} - 4 x^{2} - 4 x + 16$

Step 1

Set $x^{3} - 4 x^{2} - 4 x + 16$ equal to $0$ .

$x^{3} - 4 x^{2} - 4 x + 16 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 4 x^{2}) - 4 x + 16 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $x - 4$ .

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

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

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

Factor.

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 $b = 2$ .

$(x - 4) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x - 4) (x + 2) (x - 2) = 0$

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

$x - 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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 $(x - 4) (x + 2) (x - 2) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 4$ (Multiplicity of $1$ )

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

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

$x = 4$ (Multiplicity of $1$ )

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

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

Step 3

Identify the Zeros and Their Multiplicities x^3-4x^2-4x+16

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