Find the Roots (Zeros) f(x)=2(x-3)(x+2)^2(x-1)

$f (x) = 2 (x - 3) {(x + 2)}^{2} (x - 1)$

Set $2 (x - 3) {(x + 2)}^{2} (x - 1)$ equal to $0$ .

$2 (x - 3) {(x + 2)}^{2} (x - 1) = 0$

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 $0$ .

$x - 3 = 0$

${(x + 2)}^{2} = 0$

$x - 1 = 0$

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 ${(x + 2)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 2)}^{2}$ equal to $0$ .

${(x + 2)}^{2} = 0$

Solve ${(x + 2)}^{2} = 0$ for $x$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 3, - 2, 1$