Solve by Factoring 2x^4-72x^2=0

$2 x^{4} - 72 x^{2} = 0$

Step 1

Factor $2 x^{2}$ out of $2 x^{4} - 72 x^{2}$ .

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

$2 x^{2} (x^{2}) - 72 x^{2} = 0$

Factor $2 x^{2}$ out of $- 72 x^{2}$ .

$2 x^{2} (x^{2}) + 2 x^{2} (- 36) = 0$

Factor $2 x^{2}$ out of $2 x^{2} (x^{2}) + 2 x^{2} (- 36)$ .

$2 x^{2} (x^{2} - 36) = 0$

Step 2

Rewrite $36$ as $6^{2}$ .

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

Step 3

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 $b = 6$ .

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

Remove unnecessary parentheses.

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

Step 4

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

$x^{2} = 0$

$x + 6 = 0$

$x - 6 = 0$

Step 5

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 specified 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 6

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 8

The final solution is all the values that make $2 x^{2} (x + 6) (x - 6) = 0$ true.

$x = 0, - 6, 6$

Solve by Factoring 2x^4-72x^2=0

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