Find the Roots (Zeros) y=3(x+4)(x+1)(x-3)

$y = 3 (x + 4) (x + 1) (x - 3)$

Set $3 (x + 4) (x + 1) (x - 3)$ equal to $0$ .

$3 (x + 4) (x + 1) (x - 3) = 0$

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 $0$ .

$x + 4 = 0$

$x + 1 = 0$

$x - 3 = 0$

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 $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $3 (x + 4) (x + 1) (x - 3) = 0$ true.

$x = - 4, - 1, 3$