Find the Roots (Zeros) -2x^4+4x^3-6x^2+12x

$- 2 x^{4} + 4 x^{3} - 6 x^{2} + 12 x$

Step 1

Set $- 2 x^{4} + 4 x^{3} - 6 x^{2} + 12 x$ equal to $0$ .

$- 2 x^{4} + 4 x^{3} - 6 x^{2} + 12 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x$ out of $- 2 x^{4} + 4 x^{3} - 6 x^{2} + 12 x$ .

Factor $2 x$ out of $- 2 x^{4}$ .

$2 x (- x^{3}) + 4 x^{3} - 6 x^{2} + 12 x = 0$

Factor $2 x$ out of $4 x^{3}$ .

$2 x (- x^{3}) + 2 x (2 x^{2}) - 6 x^{2} + 12 x = 0$

Factor $2 x$ out of $- 6 x^{2}$ .

$2 x (- x^{3}) + 2 x (2 x^{2}) + 2 x (- 3 x) + 12 x = 0$

Factor $2 x$ out of $12 x$ .

$2 x (- x^{3}) + 2 x (2 x^{2}) + 2 x (- 3 x) + 2 x (6) = 0$

Factor $2 x$ out of $2 x (- x^{3}) + 2 x (2 x^{2})$ .

$2 x (- x^{3} + 2 x^{2}) + 2 x (- 3 x) + 2 x (6) = 0$

Factor $2 x$ out of $2 x (- x^{3} + 2 x^{2}) + 2 x (- 3 x)$ .

$2 x (- x^{3} + 2 x^{2} - 3 x) + 2 x (6) = 0$

Factor $2 x$ out of $2 x (- x^{3} + 2 x^{2} - 3 x) + 2 x (6)$ .

$2 x (- x^{3} + 2 x^{2} - 3 x + 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 x ((- x^{3} + 2 x^{2}) - 3 x + 6) = 0$

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

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

Factor.

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

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

Remove unnecessary parentheses.

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

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

$x = 0$

$- x + 2 = 0$

$x^{2} + 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$- x + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$- x = - 2$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

Set $x^{2} + 3$ equal to $0$ and solve for $x$ .

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

$x^{2} + 3 = 0$

Solve $x^{2} + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 3}$

Simplify $\pm \sqrt{- 3}$ .

Rewrite $- 3$ as $- 1 (3)$ .

$x = \pm \sqrt{- 1 (3)}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i \sqrt{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i \sqrt{3}, - i \sqrt{3}$

The final solution is all the values that make $2 x (- x + 2) (x^{2} + 3) = 0$ true.

$x = 0, 2, i \sqrt{3}, - i \sqrt{3}$

Step 3

Find the Roots (Zeros) -2x^4+4x^3-6x^2+12x

Related Questions

Support LCMGCF.com by disabling your adblocker.