Solve by Factoring x^3-49x=0

$x^{3} - 49 x = 0$

Step 1

Factor $x$ out of $x^{3} - 49 x$ .

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

$x \cdot x^{2} - 49 x = 0$

Factor $x$ out of $- 49 x$ .

$x \cdot x^{2} + x \cdot - 49 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 49$ .

$x (x^{2} - 49) = 0$

Step 2

Rewrite $49$ as $7^{2}$ .

$x (x^{2} - 7^{2}) = 0$

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 $b = 7$ .

$x ((x + 7) (x - 7)) = 0$

Remove unnecessary parentheses.

$x (x + 7) (x - 7) = 0$

Step 4

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

$x = 0$

$x + 7 = 0$

$x - 7 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 7

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 8

The final solution is all the values that make $x (x + 7) (x - 7) = 0$ true.

$x = 0, - 7, 7$