Find the Roots (Zeros) y=(x+1)(x-3)(x-6)

$y = (x + 1) (x - 3) (x - 6)$

Step 1

Set $(x + 1) (x - 3) (x - 6)$ equal to $0$ .

$(x + 1) (x - 3) (x - 6) = 0$

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 $0$ .

$x + 1 = 0$

$x - 3 = 0$

$x - 6 = 0$

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.

$x = 3$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $(x + 1) (x - 3) (x - 6) = 0$ true.

$x = - 1, 3, 6$

Step 3