Solve by Factoring 5x^3-20x=0

$5 x^{3} - 20 x = 0$

Step 1

Factor $5 x$ out of $5 x^{3} - 20 x$ .

Factor $5 x$ out of $5 x^{3}$ .

$5 x (x^{2}) - 20 x = 0$

Factor $5 x$ out of $- 20 x$ .

$5 x (x^{2}) + 5 x (- 4) = 0$

Factor $5 x$ out of $5 x (x^{2}) + 5 x (- 4)$ .

$5 x (x^{2} - 4) = 0$

Step 2

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

$5 x (x^{2} - 2^{2}) = 0$

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 $b = 2$ .

$5 x ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$5 x (x + 2) (x - 2) = 0$

Step 4

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

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 7

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 8

The final solution is all the values that make $5 x (x + 2) (x - 2) = 0$ true.

$x = 0, - 2, 2$