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

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

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

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

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 $0$ .

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

$x + 3 = 0$

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

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 $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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.

$x = - 1$

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

$x = 2, - 3, - 1$