Find the Roots (Zeros) x^4-2x^3-2x^2=0

$x^{4} - 2 x^{3} - 2 x^{2} = 0$

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

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

$x^{2} x^{2} - 2 x^{3} - 2 x^{2} = 0$

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

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

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

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

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

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

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

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

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

$x^{2} = 0$

$x^{2} - 2 x - 2 = 0$

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

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

$x^{2} - 2 x - 2 = 0$

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 + 8}}{2 \cdot 1}$

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{3}}{2}$ .

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

The final answer is the combination of both solutions.

$x = 1 + \sqrt{3}, 1 - \sqrt{3}$

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

$x = 0, 1 + \sqrt{3}, 1 - \sqrt{3}$

The result can be shown in multiple forms.

Exact Form:

$x = 0, 1 + \sqrt{3}, 1 - \sqrt{3}$

Decimal Form:

$x = 0, 2.73205080 \dots, - 0.73205080 \dots$

Find the Roots (Zeros) x^4-2x^3-2x^2=0

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