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

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

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

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

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 $0$ .

$x - 2 = 0$

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

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

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 ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

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

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

Solve ${(x + 1)}^{2} = 0$ for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

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

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ 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 $(x - 2) {(x + 1)}^{2} {(x - 1)}^{2} = 0$ true.

$x = 2, - 1, 1$