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

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

Step 1

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

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

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 $0$ .

$x + 1 = 0$

$x - 3 = 0$

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

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.

$x = - 1$

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

$x = - 1, 3, 1$

Step 3