Solve by Factoring 2x^2-8=0

$2 x^{2} - 8 = 0$

Step 1

Factor $2$ out of $2 x^{2} - 8$ .

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

$2 (x^{2}) - 8 = 0$

Factor $2$ out of $- 8$ .

$2 x^{2} + 2 \cdot - 4 = 0$

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

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

Step 2

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

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

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 $b = 2$ .

$2 ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$2 (x + 2) (x - 2) = 0$

Step 4

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

$x + 2 = 0$

$x - 2 = 0$

Step 5

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

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.

$x = 2$

Step 7

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

$x = - 2, 2$