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

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

Step 1

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

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

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 $0$ .

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

$x + 2 = 0$

$x - 1 = 0$

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

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

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

Solve ${(x - 3)}^{2} = 0$ for $x$ .

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 2$ equal to $0$ and solve for $x$ .

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

$x = 3, - 2, 1$

Step 3