Find the Roots (Zeros) y=(x-8)(x+3)^2

$y = (x - 8) {(x + 3)}^{2}$

Step 1

Set $(x - 8) {(x + 3)}^{2}$ equal to $0$ .

$(x - 8) {(x + 3)}^{2} = 0$

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 $0$ .

$x - 8 = 0$

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

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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 $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 8) {(x + 3)}^{2} = 0$ true.

$x = 8, - 3$

Step 3