Find the Roots (Zeros) -3x(x+4)(3x+2)(x-9)

$- 3 x (x + 4) (3 x + 2) (x - 9)$

Set $- 3 x (x + 4) (3 x + 2) (x - 9)$ equal to $0$ .

$- 3 x (x + 4) (3 x + 2) (x - 9) = 0$

Solve for $x$ .

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

$x = 0$

$x + 4 = 0$

$3 x + 2 = 0$

$x - 9 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Solve $3 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

The final solution is all the values that make $- 3 x (x + 4) (3 x + 2) (x - 9) = 0$ true.

$x = 0, - 4, - \frac{2}{3}, 9$