Solve for x (x-2)^2(x-3)(x+1)=0

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

Step 1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

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

$x - 3 = 0$

$x + 1 = 0$

Step 2

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.

$x = 2$

Step 3

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.

$x = 3$

Step 4

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.

$x = - 1$

Step 5

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

$x = 2, 3, - 1$