Find the Roots (Zeros) 0=x^3(x+3)(x-1)^2

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

Step 1

Rewrite the equation as $x^{3} (x + 3) {(x - 1)}^{2} = 0$ .

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

Step 2

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

$x^{3} = 0$

$x + 3 = 0$

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

Step 3

Set $x^{3}$ equal to $0$ and solve for $x$ .

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Solve $x^{3} = 0$ for $x$ .

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 4

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.

$x = - 3$

Step 5

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.

$x = 1$

Step 6

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

$x = 0, - 3, 1$

Step 7