Identify the Zeros and Their Multiplicities y=4x^3-4x

$y = 4 x^{3} - 4 x$

Step 1

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

$4 x^{3} - 4 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $4 x$ out of $4 x^{3} - 4 x$ .

Factor $4 x$ out of $4 x^{3}$ .

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

Factor $4 x$ out of $- 4 x$ .

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

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 1)$ .

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

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

$4 x ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$4 x (x + 1) (x - 1) = 0$

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

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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.

$x = 1$

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

$x = 0$ (Multiplicity of $1$ )

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

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

$x = 0$ (Multiplicity of $1$ )

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

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

Step 3