Find the Roots (Zeros) (x+1)^2(x-5)=0

${(x + 1)}^{2} (x - 5) = 0$

Step 1

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

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

$x - 5 = 0$

Step 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.

$x = - 1$

Step 3

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 4

The final solution is all the values that make ${(x + 1)}^{2} (x - 5) = 0$ true.

$x = - 1, 5$

Step 5