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

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

Step 1

Add $2 x^{2}$ to both sides of the equation.

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

Step 2

Factor the left side of the equation.

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

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

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

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

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

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

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

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

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

Factor $2 x^{2}$ out of $2 x^{2} (x^{2} - 2 x) + 2 x^{2} (1)$ .

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

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$2 x^{2} (x^{2} - 2 \cdot x \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 3

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

$x^{2} = 0$

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

Step 4

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 $0$ .

$x = 0$

Step 5

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 6

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

$x = 0, 1$

Step 7

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

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