Find the Roots (Zeros) 2x^3-4x^2-14x+28

$2 x^{3} - 4 x^{2} - 14 x + 28$

Step 1

Set $2 x^{3} - 4 x^{2} - 14 x + 28$ equal to $0$ .

$2 x^{3} - 4 x^{2} - 14 x + 28 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{3} - 4 x^{2} - 14 x + 28$ .

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

$2 (x^{3}) - 4 x^{2} - 14 x + 28 = 0$

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

$2 (x^{3}) + 2 (- 2 x^{2}) - 14 x + 28 = 0$

Factor $2$ out of $- 14 x$ .

$2 (x^{3}) + 2 (- 2 x^{2}) + 2 (- 7 x) + 28 = 0$

Factor $2$ out of $28$ .

$2 x^{3} + 2 (- 2 x^{2}) + 2 (- 7 x) + 2 \cdot 14 = 0$

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

$2 (x^{3} - 2 x^{2}) + 2 (- 7 x) + 2 \cdot 14 = 0$

Factor $2$ out of $2 (x^{3} - 2 x^{2}) + 2 (- 7 x)$ .

$2 (x^{3} - 2 x^{2} - 7 x) + 2 \cdot 14 = 0$

Factor $2$ out of $2 (x^{3} - 2 x^{2} - 7 x) + 2 \cdot 14$ .

$2 (x^{3} - 2 x^{2} - 7 x + 14) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 ((x^{3} - 2 x^{2}) - 7 x + 14) = 0$

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

$2 (x^{2} (x - 2) - 7 (x - 2)) = 0$

Factor.

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

$2 ((x - 2) (x^{2} - 7)) = 0$

Remove unnecessary parentheses.

$2 (x - 2) (x^{2} - 7) = 0$

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

$x - 2 = 0$

$x^{2} - 7 = 0$

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.

$x = 2$

Set $x^{2} - 7$ equal to $0$ and solve for $x$ .

Set $x^{2} - 7$ equal to $0$ .

$x^{2} - 7 = 0$

Solve $x^{2} - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$x^{2} = 7$

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

$x = \pm \sqrt{7}$

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.

$x = \sqrt{7}$

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

$x = - \sqrt{7}$

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

$x = \sqrt{7}, - \sqrt{7}$

The final solution is all the values that make $2 (x - 2) (x^{2} - 7) = 0$ true.

$x = 2, \sqrt{7}, - \sqrt{7}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 2, \sqrt{7}, - \sqrt{7}$

Decimal Form:

$x = 2, 2.64575131 \dots, - 2.64575131 \dots$

Step 4

Find the Roots (Zeros) 2x^3-4x^2-14x+28

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