Identify the Zeros and Their Multiplicities f(x)=2x^3-x^2-128x+64

$f (x) = 2 x^{3} - x^{2} - 128 x + 64$

Step 1

Set $2 x^{3} - x^{2} - 128 x + 64$ equal to $0$ .

$2 x^{3} - x^{2} - 128 x + 64 = 0$

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.

$(2 x^{3} - x^{2}) - 128 x + 64 = 0$

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

$x^{2} (2 x - 1) - 64 (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x^{2} - 64) = 0$

Rewrite $64$ as $8^{2}$ .

$(2 x - 1) (x^{2} - 8^{2}) = 0$

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 $b = 8$ .

$(2 x - 1) ((x + 8) (x - 8)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x + 8) (x - 8) = 0$

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

$2 x - 1 = 0$

$x + 8 = 0$

$x - 8 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

The final solution is all the values that make $(2 x - 1) (x + 8) (x - 8) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = \frac{1}{2}$ (Multiplicity of $1$ )

$x = - 8$ (Multiplicity of $1$ )

$x = 8$ (Multiplicity of $1$ )

$x = \frac{1}{2}$ (Multiplicity of $1$ )

$x = - 8$ (Multiplicity of $1$ )

$x = 8$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=2x^3-x^2-128x+64

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