Find the Roots (Zeros) f(x)=2x^3-x^2-32x+16

$f (x) = 2 x^{3} - x^{2} - 32 x + 16$

Set $2 x^{3} - x^{2} - 32 x + 16$ equal to $0$ .

$2 x^{3} - x^{2} - 32 x + 16 = 0$

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.

$(2 x^{3} - x^{2}) - 32 x + 16 = 0$

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

$x^{2} (2 x - 1) - 16 (2 x - 1) = 0$

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

$(2 x - 1) (x^{2} - 16) = 0$

Rewrite $16$ as $4^{2}$ .

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

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 $b = 4$ .

$(2 x - 1) ((x + 4) (x - 4)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x + 4) (x - 4) = 0$

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

$2 x - 1 = 0$

$x + 4 = 0$

$x - 4 = 0$

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 $x + 4$ equal to $0$ and solve for $x$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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.

$x = 4$

The final solution is all the values that make $(2 x - 1) (x + 4) (x - 4) = 0$ true.

$x = \frac{1}{2}, - 4, 4$

Find the Roots (Zeros) f(x)=2x^3-x^2-32x+16

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