Find the Roots (Zeros) y=3(x-2)(x+1)(x-4)

$y = 3 (x - 2) (x + 1) (x - 4)$

Step 1

Set $3 (x - 2) (x + 1) (x - 4)$ equal to $0$ .

$3 (x - 2) (x + 1) (x - 4) = 0$

Step 2

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 $0$ .

$x - 2 = 0$

$x + 1 = 0$

$x - 4 = 0$

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.

$x = 2$

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 $x - 4$ equal to $0$ and solve for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $3 (x - 2) (x + 1) (x - 4) = 0$ true.

$x = 2, - 1, 4$

Step 3